/^  ^,(ru. 


ELEMENTS 


INTEGRAL    CALCULUS 


KEY  TO   THE  SOLUTION  OF  DIFFERENTIAL 
EQUATIONS. 


BY 

WILLIAM   ELWOOD   BYERLY,   Ph.D.,  5, 

PKOFESSOK  OF  MATHEMATICS  IN  HARVARD  UNIVERSITY. 

BOSTON  COLLEGE 
PHYSICS  DEPT. 


BOSTON: 

PUBLISHED  BY  GINN,  HEATH,   &  CO. 
1881. 


ELEMENTS 


INTEGRAL    CALCULUS 


KEY  TO   THE  SOLUTION  OF  DIFFERENTIAL 
EQUATIONS. 


BY 

WILLIAM   ELWOOD   BYERLY,   Ph.D.,  5, 

PROFESSOR  OF  MATHEMATICS  IN  HARVARD  UNIVERSITY. 

BOSTON  COLLEGE 
PHYSICS  DEPT. 


BOSTON: 

PUBLISHED  BY  GINN,   HEATH,   &  CO. 
1881. 


Enterea  according  to  Act  of  Congress,  in  the  year  1S81,  by 

William  Elavood  Byekly, 
in  the  office  of  the  Librarian  of  Congress,  at  Washington. 


GiNN,  Heath,  &  Co. 

J.  S.  CusHiNG,  Printer,  16  Hawley  Street, 

Boston. 


260413 


PREFACE. 


The  following  volume  is  a  sequel  to  my  treatise  on  the 
Differential  Calculus,  and,  like  that,  is  written  as  a  text-book. 
The  last  chapter,  however,  a  Key  to  the  Solution  of  Differential 
Equations,  maj^  prove  of  service  to  working  mathematicians. 

I  have  used  freely  the  works  of  Bertrand,  Benjamin  Peirce, 
Todlmnter,  and  Boole ;  and  I  am  much  indebted  to  Professor 
J.  M.  Peirce  for  criticisms  and  suggestions. 

I  refer  constantly  to  my  work  on  the  Differential  Calculus 
as  Volume  I. ;  and  for  the  sake  of  convenience  I  have  added 
Chapter  V.  of  that  book,  which  treats  of  Integration,  as  an 

appendix  to  the  present  volume. 

W.  E.  BYERLY. 
Cambridge,  1881. 


ANALYTICAL  TABLE  OF  CONTENTS. 


CHAPTER    I. 


SYMBOLS   OF   OPERATION. 

Article.  Page. 

1.  FivnctioViBl  symbols  regarded  as  symbols  of  ojjeration 1 

2.  Compoimd  function ;  compound  operation 1 

3.  Commutative  or  relatively  free  operations 1 

4.  Distributive  or  linear  operations 2 

5.  The  compounds  of  distributive  operations  are  distributive  ...  2 

6.  Symbolic  exponents 2 

7.  The  law  of  indices 2 

8.  The  interpretation  of  a  zero  exponent 3 

9.  The  interpretation  of  a  negative  exponent .  3 

10.   When  operations  are  commutative  and  distributive,  the  sym- 
bols which  represent  them  may  be  combined  as  if  they  were 

algebraic  quantities 3 


CHAPTER    II. 

EVIAGESTAKIES. 

11.  Usual  definition  of  an  imaginary.   Imaginaries  first  forced  upon 

our  attention  in  connection  with  quadratic  equations  ....  5 

12.  Treatment  of  imaginaries  purely  arbitrary  and  conventional  .    .  (J 

13.  V^  defined  as  a  symbol  of  operation fl 

14.  The  rules  in  accordance  with  which  the  symbol  V  —  1  is  used. 

V^  distributive  and  commutative  with  symbols  of  quantity  .  7 

15.  Interpretation  of  powers  of  V^l 7 

16.  Imaginary  roots  of  a  quadratic 8 

17.  Typical  form  of  an  imaginary 8 

18.  Geometrical  representation  of  an  imaginary.     Beals  and  2mre 

imaginaries.    An  interpretation  of  the  operation  V^^    ...      8 

19.  The  sum,  the  product,  and  the  quotient  of  two  imaginaries, 

a  +  b  V^  and  c  +  d  V^,   are  imaginaries  of  the  typical 
form 10 


vi  INTEGRAL   CALCULUS. 

Article.  Page. 

20.  Second  typical  form  r  (cos  f  +  V— 1  sin  (p).     Modulus  and  argii- 

ment.     Examples 10 

21.  The  modulus  of  the  sum  of  two  imaginaries  is  never  greater 

than  the  sum  of  their  moduli 11 

22.  Modulus  and  argument  of  the  product  of  imaginaries 12 

23.  Modulus  and  argument  of  the  quotient  of  two  imaginaries     .   .  13 

24.  Modulus  and  argument  of  a  power  of  an  imaginary 13 

25.  Modulus  and  argument  of  a  root  of  an  imaginary.     Example    .  14 

26.  Relation  between  the  n  nth  roots  of  a  real  or  an  imaginary   ,    .  14 

27.  The  imaginary  roots  of  1  and  —  1.     Examples 15 

28.  Conjugate  imaginaries.     Examples 17 

29.  Transcendental  functions  of  an  imaginary  variable  best  defined 

by  the  aid  of  series 17 

30.  Couvergency  of  a  series  containing  imaginary  terms 18 

31.  Exponential  functions  of  an  imaginary.     Definition  of  e"  where 

z  is  imaginary 19 

32.  The  law  of  indices  holds  for  imaginary  exponentials.   Example  20 

33.  Logarithmic  functions  of  an  imaginary.      Definition  of  log  z. 

Log  2;  a  pe?'i;ofZtc  function.     Example 21 

34.  Trigonometric  functions  of  an  imaginary.     Definition  of  sin  z 

and  cos  z.    Example 22 

35.  Sin  z  and  cos  z  expressed  in  exponential  form.    The  fundamen- 

tal formulas  of  Trigonometry  hold  for  imaginaries  as  well  as 

for  reals.     Examples 22 

36.  Differentiation  of  Functions  of  Imaginary  Variables.     The  de- 

rivative of  a  function  of  an  imaginary  is  in  general  indetermi- 
nate    24 

37.  In  difi"erentiating.  we  may  treat  the  V—i  like  a  constant  factor. 

Example.     Two  forms  of  the  differential  of  the  independent 

variable , 24 

38.  Differentiation  of  a  power  of  z.     Example 25 

39.  Differentiation  of  e^.     Example 26 

40.  Differentiation  of  log  z 26 

41.  Diflerentiatiou  of  sir^g  and  coss 26 

42.  Formulas  for  direct  integration  (I.,  Art.  74)  hold  when  x  is 

imaginary 27 

43.  Hyperholic  Functions 27 

44.  Examples.     Properties  of  Hyperbolic  Functions 28 

45.  Differentiation  of  Hyperbolic  Functions 28. 

46.  Anti-hyperbolic  functions.     Examples 28 

47.  Anti-hyperbolic  functions  expressed  as  logarithms 29 

48.  Formulas  for  the  direct  integration  of  some  irrational  forms    .  30 


TABLE  OF  CONTENTS.  vii 

CHAPTER  III. 

GENERAL  METHODS  OF  INTEGRATINCr. 

Article.  Page. 

49.  Integral  regarded  as  the  inverse  of  a  differential 32 

50.  If /x  is  any  function  whatever  of  x,  fx.dx  has  an  integral,  and 

but  one,  except  for  the  presence  of  an  arbitrary  constant  .    .    32 

51.  A  definite  integral  contains  no  arbitrary  constant,  and  is  a  func- 

tion of  the  values  between  which  the  sum  is  taken.     Exam- 
ples    33 

52.  Definite  integral  of  a  disco  Hf Ml i(o?fS  function 33 

53.  Formulas  for  direct  integration 34 

54.  Integration  by  substitution.     Examples 36 

55.  Integration  by  parts.    Examples.     Miscellaneous  examples  in 

integration 37 


CHAPTER    IV. 

EATIONAJL  FRACTIONS. 

56.  Integration  of  a  rational  algebraic  polynomial.    Rational  frac- 

tions, proper  and  improper 40 

57.  Every  proper  rational  fraction  can  be  reduced  to  a  sum  of  sim- 

pler fractions  with  constant  numerators 40 

58.  Determination  of  the  numerators  of  the  partial  fractions  by 

indirect  methods.     Examples 42 

59.  Direct  determination  of  the  numerators  of  the  partial  fractions  43 

60.  Illustrative  examples 45 

61.  Illustrative  example 46 

62.  Integration  of  the  partial  fractions 48 

63.  Treatment  of  imaginary  values  which  may  occur  in  the  partial 

fractions.    Examples 49 


CHAPTER   V. 

REDUCTION  FORMULAS. 

64.  Formulas  for  raising  or  lowering  the  exponents  in  the  form 

x'^-'^{a-\-hx'^)Pdx 52 

65.  Consideration  of  special  cases.    Examples 54 


Viii  USTTEGEAL   CALCULUS. 

CHAPTER    VL 

IRRATIONAL   FORMS. 
Article.  ^^Se 

66.  lategration  of  the  form  f(x,  Va  +  bx)dx.    Examples 56 

67.  lutegrationof  the  form/(x,  Vc+ \/a  + to)f?x.     Examples    .    .  57 

68.  Integration  of  the  form /(x,  Va  +  ta  +  cx'Of^a; 57 

69.  Illustrative  example.     Examples 59 

70.  Integration  of  the  form /fx,A/^^-t-V^3;.    Example 61 

\       ^Ix+mJ 

71.  Application  of  the  Beduction  Formulas  of  Chapter  V.  to  irra- 

tional forms.     Examples 61 

72.  A  function  rendered  irrational  through  the  presence  under 

the  radical  sign  of  a  polynomial  of  higher  degree  than  the 
second  cannot  ordinarily  be  integrated.     Elliptic  Integrals   .    62 


CHAPTER    VII. 

TRANSCENDENTAL  FUNCTIONS. . 

73.  Use  of  the  method  of  Integration  by  Farts.    Examples     ....  63 

74.  Reduction  Formulas  for  sin"  X  and  cos"  X.     Examples 64 

75.  Integration  of  (sin-ix)«(:?x.     Examples 65 

76.  Use  of  the  method  of  Integration  by  Substitution 66 

77.  Integration  of  sin™  X  COS" x.dx.    Examples 67 


CHAPTER  VIII. 

r)EFIN^TE   INTEGRALS. 

78.  Computation  of  a  definite  integral  as  the  limit  of  a  sum  ....    69 

79.  Illustrative  example.     Examples 70 

80.  Usual  method  of  obtaining  the  value  of  a  definite  integral. 

Examples • 70 

81.  Application  of  reduction  formulas  to  definite  integrals.     Ex- 

amples      71 

82.  The  Gamma  Function 73 

83.  The  x>rincipal  value  of  the  definite  integral  of  a  discontinuous 

function.     Example 74 

84.  Difi'erentiation  and  integration  of  a  definite  integral  when  the 

limits  of  integration  are  constant 74 


TABLE   OF   CONTENTS.  IX 

Article.  Page. 

85.  Simplification  of  a  complicated  form  by  differentiation  under 

the  sign  of  definite  integration.     Example 76 

86.  Simplification  of  a  complicated  form  by  integration  under  the 

sign  of  definite  integration.     Examples 77 

87.  Differentiation  of  a  definite  integral  when  the  limits  of  integra- 

tion are  not  constant.     Example 77 

88.  Definite  integrals  taken  between  imaginary  limits 78 


CHAPTEK  IX. 

LENGTHS  OF  CURVES. 

89.  Formulas  for  sin  t  and  cos  t  in  terms  of  the  length  of  the  arc  .  79 

90.  The  equation  of  the  Catenary  obtained.     Example 79 

91.  The  equation  of  the  Tractrix.     Examples 81 

92.  Length  of  an  arc  in  rectangular  coordinates 83 

93.  Length  of  the  arc  of  the  C?/do«cL     Example 84 

94.  Another  method  of  rectifying  the  Cycloid 85 

95.  Hecti&ca.tion  of  the  Epicycloid.     Examples 85 

96.  Arc  of  the  Ellipse.     Auxiliary  angle.     Example 86 

97.  Length  of  an  arc.     Polar  coordinates  . 87 

98.  Equation  of  the  Logarithmic  Spiral 87 

99.  Eectification  of  the  Logarithmic  Spiral.     Examples 88 

100.  Rectification  of  the  Cardioide 88 

101.  Involutes.     Illustrative  example.     Example 89 

102.  The  involute  of  the  Cycloid.     Example 91 

103.  Intrinsic  equation  of  a  curve.     Example 92 

104.  Intrinsic  equation  of  the  Epicycloid.     Example 93 

105.  Intrinsic  equation  of  the  Logarithmic  Spiral 94 

106.  Method  of  obtaining  the  intrinsic  equation  from  the  equation 

in  rectangular  coordinates.    Examples 94 

107.  Intrinsic  equation  of  an  evolute 96 

108.  Illustrative  examples.     Examples 96 

109.  The  evolute  of  an  Epicycloid.    Example 97 

110.  The  intrinsic  equation  of  an  involute.   Illustrative  examples  .  9§ 

111.  Limiting  form  approached  by  an  involute  of  an  involute  .    .    .  99 

112.  Method  of  obtaining  the  equation  in  rectangular  coordinates 

from  the  intrinsic  equation.     Illustrative  example 100 

113.  Rectification  of  Curves  in  Space.    Examples 101 


INTEGEAL    CALCULUS. 


CHAPTER  X. 

AREAS. 
Article.  Page. 

114.  Areas  expressed  as  definite    integrals,   rectangular  coordi- 

nates.    Examples 103 

115.  Areas  expressed  as  definite  integrals,  polar  coordinates   .    .    .104 

116.  Area  between  the  catenary  and  the  axis 104 

117.  Area  between  the  tractrix  and  the  axis.     Example 104 

118.  Area  between  a  curve  and  its  asymptote.     Examples     ....  105 

119.  Area  of  circle  obtained  by  the  aid  of  an  auxiliary  angle.    Ex- 

amples     lOG 

120.  Area  between  two  curves  (rect.  coor.).     Examples 107 

121.  Areas  in  Polar  Coordinates.     Examples 108 

122.  Problems  in  areas  can  often  be  simplified  by  transformation 

of  coordinates.    Examples Ill 

123.  Area  between  a  curve  and  its  evolute."    Examples Ill 

124.  Holditch's  Theorem.    Examples 112 

125.  Areas  by  a  double  integration  (rect.  coor.)      114 

126.  Illustrative  examples.    Examples 115 

127.  Areas  by  a  double  integration  (polar  coor.).    Example  ....  116 

CHAPTER   XI. 

AREAS    OF   SURFACES. 

128.  Area  of  a  surface  of  revolution  (rect.  coor.).     Example  ....  118 

129.  Illustrative  examples.     Examples 119 

180.   Area  of  a  surface  of  revolution  by  transformation  of  coordi- 
nates.    Example 120 

131.  Kvedi  of  a,  surface  of  revolution  (j^o\&Y  coor.').     Examples  .   .    .122 

132.  Area  of  any  siirface  by  a  double  integration 122 

133.  Illustrative  example.     Examples 125 

^CHAPTER    XIL 

VOLUMES. 

134.  Volume  by  a  single  integration.     Example 128 

135.  Volume  of  a  conoid.     Examples 129 

136.  Volume  of  an  ellipsoid.     Examples 130 

137.  Volume  of  a  solid  of  revolution.     Single  integration.     Exam- 

ples   131 

138.  Volume  of  a  solid  of  revolution.     Double  integration.     Exam- 

ples   132 


TABLE    OF    CONTENTS.  xi 

Article.  Page. 

139.  Yolnme  of  a  solid  of  revolution.    Polar  formula.     Example.    .134 

140.  Volume  of  any  solid.     Triple  iutegratiou.     Rectangular  coor- 

dinates.    Examples 135 

141.  Volume  of  miy  solid.     Triple  integration.     Polar  coordinates. 

Example 138 

CHAPTER    XIII. 

CENTRES   OF  GRAVITY. 

142.  Centre  of  Gravity  defined 139 

143.  General  formulas  for  the  coordinates  of  the  Centre  of  Gravity 

of  any  mass.     Example 139 

144.  Centre  of  Gravity  of  a  homogeneous  body 141 

145.  Centre  of  Gravity  of  a  j9Zrt we  area.     Examples 141 

146.  Centre  of  Gravity  of  a  homogeneous  solid  of  revolution.     Ex- 

amples     • 144 

147.  Centre  of  Gravity  of  an  arc  ;  of  a  surface  of  revolution.    Exam- 

ples   146 

148.  Properties  of  Guldin.    Examples 147 

CHAPTER    XIV. 

MEAN  VALUE  AND   PROBABILITY. 

149.  References 149 

150.  Mean  value  of  a  contimiously  varying  quantity.     The  mean  dis- 

tance of  all  the  points  of  the  circumferences  of  a  circle  from 
a  fixed  point  on  the  circumference.  The  mean  distance  of 
points  on  the  surface  of  a  circle  from  a  fixed  point  on  the 
cii'cumference.  The  mea^i  distance  between  two  points 
within  a  given  circle 149 

151.  Problems  in  the  application  of  the  Integral  Calculus  to  pro&a- 

bilities.     Random  straight  lines.     Examples 151 

CHAPTER  XV. 

KEY  TO   THE   SOLUTION   OF   DIFFERENTIAL  EQUATIONS. 

152.  Description  of  Key 158 

153.  Definition   of  the  terms   differential  equation,  order,  degree, 

linear,  general  sohition  or  complete  primitive,  singular  solu- 
tion. 

154.  Examples  illustrating  the  use  of  tfie  Key 159 

Key 1G3 

Examples  under  Key 180 


INTEGEAL    CALCULUS. 


CHAPTER    I. 

SYMBOLS    OF    OPEKATIOIST. 


1.  It  is  often  convenient  to  regard  a  functional  symbol  as 
indicating  an  operation  to  he  ])er formed  upon  the  expression 
tvhich  is  ivritten  after  the  symbol.  From  this  point  of  view  the 
s^Tnbol  is  called  a  symbol  of  operation,  and  the  expression  writ- 
ten after  the  s3'mbol  is  called  the  subject  of  the  operation. 

Thus  the  S3'mbol  D^  in  D^{x-y)  indicates  that  the  operation  of 
differentiating  with  respect  to  a;  is  to  be  performed  upon  the 
subject  (x^y). 

2.  If  the  result  of  one  operation  is  taken  as  the  subject  of  a 
second,  there  is  formed  what  is  called  a  compound  function. 

Thus  log  sin?;  is  a  compound  function,  and  we  may  speak  of 
the  taking  of  the  log  sin  as  a  compound  operation. 

3.  When  two  operations  are  so  related  that  the  compound 
operation,  in  which  the  result  of  performing  the  first  on  any 
subject  is  taken  as  the  subject  of  the  second,  leads  to  the  same 
result  as  the  compound  operation,  in  which  the  result  of  per- 
forming the  second  on  the  same  subject  is  taken  as  the  subject 
of  the  first,  the  two  operations  are  commutative  or  relatively  free. 

Or  to  formulate  ;  if 

fFu^Ffu, 

the  operations  indicated  by  /  and  F  are  commutative. 


2  INTEGRAL  CALCULUS.  [Art.  4. 

For  example  ;  the  operations  of  partial  differentiation  with 

respect  to  two  independent  variables  x  and  y  are  commutative, 

for  we  know  that 

D,DyU  =  DyD,u.  (I.  Art.  197). 

The  operations  of  taking  the  sine  and  of  taking  the  logarithm 
are  not  commutative,  for  log  sin  w  is  not  equal  to  sin  log  m. 

4.  If  f{u±v)=fu±fv 

where  u  and  v  are  any  subjects,  the  operation  /is  distributive  or 
linear. 

The  operation  indicated  b}'  d  and  the  operation  indicated  by 
D^  are  distributive,  for  we  know  that    . 

d{u  ±v)  =  du  ±dv, 

and  that  D^{u  ±v)  =  D^u  ±  D^v. 

The  operation  sin  is  not  distributive,  for  sin(i(H-'y)  is  not 
equal  to  sinu  +  sin-y. 

5.  The  com^jounds  of  distributive  operations  are  distributive.  ■ 
Let  /  and  F  indicate  distributive  operations,  then /i^  will  be 

distributive ;  for 

F{u  ±v)  =  Fu  ±  Fv, 
therefore       fF{u  ±v)=  f{Fu  ±  Fv)  =  fFu  ±  fFv. 

6.  The  repetition  of  any  operation  is  indicated  by  writing  an 
exponent.,  equal  to  the  number  of  times  the  op>eratio7i  is  ^je?-- 
formed^  after  the  symbol  of  the  operation. 

Thus  log^a;  means  log  log  log  .'c  ;  d^u  means  dddui 

In  the  single  case  of  the  trigonometric  functions  a  different 

use  of  the  exponent  is  sanctioned  by  custom,  and  sin^u  means 

(sin^t)^  and  not  sin  sinw. 

7.  If  m  and  n  are  ivhole  numbers  it  is  easily  proved  that 


Chap.  I.]  SYMBOLS   OF   OPERATION.  3 

This  formula  is  assumed  for  all  values  of  m  and  n,  and  nega- 
tive and  fractional  exponents  are  interpreted  by  its  aid.  It  is 
called  the  laio  of  indices. 

8.  To  find  what  interpretation  must  be  given  to  a  zero  ex- 
^  '  m  =  0         in  the  formula  of  Art.  7. 

/o/«t^=/o +  »'«=/"«, 

or,  denoting /""it  hj  v,  f°v  =  v. 

That  is  ;  a  symbol  of  operation  with  the  exponent  zero  has  no 
effect  on  the  subject,  and  may  be  regarded  as  multiplying  it  b}' 
unit}'. 

9.  To  interpret  a  negative  exponent,  let 

m=  —n        in  the  formula  of  Art.  7. 

/-  ''f"-u^f-''  +  ''u=fu  =  u. 

If  we  call  f"u  =  V,  then  f-''v  =  u. 

If  II  =  1 

we  get  f~^fu=ic, 

and  the  exponent  —1  indicates  what  we  have  called  the  anti- 
function  of  fu.     (I.    Art.  72.) 

The  exponent  —  1  is  used  in  this  sense  even  with  trigonometric 
functions. 

10.  "Wlien  two  operations  are  commutative  and  distributive, 
the  symbols  which  represent  them  ma}'  be  combined  precisely  as 
if  they  were  algebraic  quantities. 

For  they  obey  the  laws, 

a  (m  -j-  n)  =  am  +  an, 

am  =  ma, 

on  which  all  the  operations  of  arithmetic  and  algebra  are  founded. 


4  INTEGEAL,  CALCFLUS.  [Art.  10. 

For  example ;  if  the  operation  {D^  +  Dy)  is  to  be*  performed 
n  times  in  succession  on  a  subject  tt,  we  can  expand  (Z>^  +  Z)j,)'* 
precisel}'  as  if  it  were  a  binominal,  and  tlien  perform  on  u  the 
operations  indicated  by  tlie  expanded  expression. 


Chap.  II.]  mAGIXAKIES. 


CHAPTER    II. 

IMAGINAEIES. 

11.  An  imaginary  is  usually  defined  in  algebra  as  the  indi- 
cated even  root  of  a  negative  quantity,  and  although  it  is  clear 
that  there  can  he  no  quantity  that  raised  to  an  even  power  will 
be  negative,  the  assumption  is  made  that  an  imaginar}'  can  be 
treated  Wko.  any  algebraic  quantit}'. 

Imaginaries  are  first  forced  upon  our  notice  in  connection 

with  the  subject  of  quadratic  equations.    Considering  the  t3'pical 

quadratic  „ 

X'  -\- ax -\- 0  =  0, 

we  find  that  it  has  two  roots,  and  that  these  roots  possess  cer- 
tain important  properties.  For  example  ;  their  sum  is  —a  and 
their  product  is  h.  "We  are  led  to  the  conclusion  that  every 
quadratic  has  two  roots  whose  sum  and  whose  product  are 
simply  related  to  the  coefficients  of  the  equation. 

On  trial,  however,  we  find  that  there  are  quadratics  having 
but  one  root,  and  quadratics  having  no  root. 

For  example  ;  if  we  solve  the  equation 

cc2-2a;  +  l  =  0, 

we  find  that  the  onl}'  value  of  x  which  will  satisfy  it  is  unity ; 
and  if  we  attempt  to  solve  ' 

x'  —  2x  +  2  =  0, 

we  find  that  there  is  no  value  of  x  which  will  satisfy-  the  equation. 

As  these  results  are  apparently  inconsistent  with  the  conclu- 
sion to  which  we  were  led  on  soMng  the  general  equation,  we 
naturall}'  endeavor  to  reconcile  them  with  it. 

The  difficulty  in  the  case  of  the  equation  which  has  but  one 


6  INTEGRAL   CALCULUS.  [Art.  12. 

root  is  easily  overcome  by  regarding  it  as  having  two  equal  roots. 
Thus  we  can  saj*  that  each  of  the  two  roots  of  the  equation 

ic^  -  2  .T  +  1  =  0 

is  equal  to  1  ;  and  there  is  a  decided  advantage  in  looking  at  the 
question  from  this  point  of  view,  for  the  roots  of  tliis  equation 
will  possess  the  same  properties  as  those  of  a  quadratic  having 
unequal  roots.  The  sum  of  the  roots  1  and  1  is  minus  the  co- 
efficient of  X  in  the  equation,  and  then'  product  is  the  constant 
term. 

To  overcome  the  difficulty  presented  by  the  equation  which 
has  no  root  we  are  driven  to  the  conception  of  imaginaries. 

12.  An  imaginary  is  not  a  quantity,  and  the  treatment  of 
imaginaries  is  purely  arbitrary  and  conventional.  We  begin  by 
lading  down  a  few  arbitrar}'^  rules  for  our  imaginary  expressions 
to  obe}',  which  must  not  involve  an}'  contradiction ;  and  we 
must  perform  all  our  operations  upon  imaginaries,  and  must 
interpret  all  our  results  by  the  aid  of  these  rules. 

Since  imaginaries  occur  as  roots  of  equations,  thej^  bear  a  close 
analog}'  with  ordinary  algebraic  quantities,  and  they  have  to  be 
subjected  to  the  same  opei'ations  as  ordinary  quantities  ;  thei'e- 
fore  our  rules  ought  to  be  so  chosen  that  the  results  may  be 
comparable  with  the  results  obtained  when  we  are  dealing  with 
real  quantities. 

13.  By  adopting  the  convention  that 

V—  a^  =  a  V  — 1, 

where  a  is  supposed  to  be  real,  we  can  reduce  all  our  imaginary 
algebraic  expi-essions  to  forms  where  V  —  1  is  the  only  peculiar 
symbol.  This  s}'mbol  V  — 1  we  shall  define  and  use  as  the  sym- 
bol of  some  operation,  at  present  unknown,  the  repetition  ofivhich 
has  the  effect  of  changing  the  sign  of  the  subject  of  the  operation. 
Thus  in  a'\l  —\  the  symbol  V  —  1  indicates  that  an  operation 
is  performed  upon  a  which,  if  repeated,  will  change  the  sign 

of  o.     That  is,  

a(V  — 1)^=  —a. 


Chap.  II.]  IMAGINAEIES.  7 

From  this  point  of  view  it  would  be  more  natural  to  write  the 
symbol  before  instead  of  after  the  subject  on  which  it  operates, 
(V  — l)a  instead  of  aV  — 1,  and  this  is  sometimes  done;  but 
as  the  usage  of  mathematicians  i^  overwhelmingi}^  in  favor  of  the 
second  form,  we  shall  emplo}'  it,  merel}^  as  a  matter  of  con- 
venience, and  remembering  that  a  is  the  subject  and  the  V  — 1 
the  symbol  of  operation. 

14.  The  rules  in  accordance  with  which  we  shall  use  our  new 
s^'mbol  are,  first, 

In  other  words,  the  operation  indicated  by  V  — 1  is  to  be  dis- 
tributive (Art.  4)  ;   and  second, 

aV"^=(V^l)a,  [2] 

or  our  sj'mbol  is  to  be  commutative  with  the  sj'mbols  of  quantity 
(Art.  3) . 

These  two  conventions  will  enable  us  to  use  our  S3'mbol  in 
algebraic  operations  precisel}'  as  if  it  were  a  quantity  (Art.  10). 

When  no  coefficient  is  written  before  V  —  1  the  coefficient  1 
will  be  understood,  or  unity  will  be  regarded  as  the  subject  of 
the  operation. 

15.  Let  us  see  what  interpretation  we  can  get  for  powers  of 
V  — 1  ;  that  is,  for  repetitions  of  the  operation  indicated  by  the 
symbol. 

(^^1)0=1  (Art.  8), 

(V^l)i=V^3, 

(  V^l ) '  =  - 1 ,  by  definition  (Art.  13), 

(V^ri)3=  (V31)2V^ri  =  -V^I,     by  definition, 
(V31)4=_(V^1)2        =1, 

(V^l)5=iV^l  =v^=n, 

(V^r-i)6^(V^^)^  =-1, 

and  so  on,  the  values  V— 1,   —1,    —  V—  1,   1,  occumng  in 


8  INTEGRAL  CALCULUS.  [Akt.  1G. 

cj'cles  of  four.     "We  can  formulate  this  as  follows  ;  let  n  be  zero 
or  anj'^  positive  whole  number,  then, 


16.  The  definition  we  have  given  for  the  square  root  of  a 
negative  quantity,  and  the  rules  we  have  adopted  concerning  its 
use,  enable  us  to  remove  entirel}^  the  difficult}'  felt  in  dealing 
with  a  quadratic  which  does  not  have  real  roots.  Take  the 
equation 

a;2_2a;  +  5  =  0.  (1) 

Solving  by  the  usual  method,  we  get 

x=l  ±  V^^ ; 

V^^  =  27"^,  by  Art.  13  [1]  ; 

hence  iK=  1  +  2V^  or  1  —  2 V"^. 

On  substituting  these  results  in  turn  in  the  equation  (1),  per- 
forming the  operations  hj  the  aid  of  our  conventions  (Art.  14 
[1]  and {2]),  ancl  interpreting  (V  — l)^  b}'  Art.  15,  we  find  that 
they  both  satisfy  the  equation,  and  that  they  can  therefore  be 
regarded  as  entirel}-  analogous  to  real  roots.  "We  find,  too,  that 
their  sum  is  2  and  that  their  product  is  5,  and  consequentl}'  that 
thej'  bear  the  same  relations  to  the  coefficients  of  the  equation  as 
real  roots. 

17.  An  imaginary  root  of  a  quadratic  can  alwaj's  be  reduced 
to  the  form  a-j-b  V— 1  where  a  and  b  are  real,  and  this  is  taken 
as  the  general  type  of  an  imaginary  ;  and  part  of  our  work  will 
be  to  show  that  when  We  subjecfimaginaries  to  .the  ordinary 
functional  operations,  all  our  results  are  reducible  to  this  tj'pical 
form. 


Chap.  II.] 


IMAGINAEIES. 


9 


18.  We  have  defined  V  —  1  as  the  symbol  of  an  operation 
whose  repetition  changes  the  sign  of  the  subject. 

Several  different  interpretations  of  this  operation  have  been 
suggested,  and  the  following  one,  in  which  every  imaginary  is 
graphicall}"  represented  by  the  position  of  a  point  in  a  plane,  is 
commonly  adopted,  and  is  found  exceedingly  useful  in  suggest- 
ing and  interpreting  relations  between  different  imaginaries  and 
between  imaginaries  and  reals. 

In  the  Calculus  of  Imaginaries,  a  +  &  V  —  1  is  taken  as  the 
general  sj'mbol  of  quantity.  If  b  is  equal  to  zero,  a-\-b  V  — 1 
reduces  to  a,  and  is  real;  if  a  is  equal  to  zero,  a  +  &  V  —  1  re- 
duces to  b  V  — 1 ,  and  is  called  a  piire  imaginary. 

a  +  5  V  —  1  is  represented  by  the  position  of  a  point  referred 
to  a  pair  of  rectangular  axes,  as  in  analytic  geometry,  a  being 
taken  as  the  abscissa  of  the 
point  and  b  as  its  ordinate. 
Thus  in  the  figure  the  position 
of  the  point  P  represents  the 
imaginary  a-\-b  V— 1 . 

If  6  =  0. 

and  our  quantity'  is  real,  P  will 

lie  on  the  axis  of  X,  which  on 

that  account  is  called  the  axis  of 

reals;  if 

a  =  0, 

and  we  have  a  j^ure  imaginary,  P  will  lie  on  the  axis  of  Y, 
which  is  called  the  axis  of  pure  imaginaries. 

Since  a  and  a  V  — 1  are  represented  by  points  equally  distant 
from  the  origin,  and  lying  on  the  axis  of  reals  and  the  axis  of 
pure  imaginaries  respectively,  we  maj^  regard  the  operation 
indicated  by  V  — 1  as  causing  the  point  representing  the  subject 
of  the  operation  to  rotate  about  the  origin  through  an  angle  of 
90°.  A  repetition  of  the  operation  ought  to  cause  the  point  to 
rotate  90°  further,  and  it  does  ;  for 

a(V— 1)^=  —a, 
and  is  represented  b}^  a  point  at  the  same  distance  from  the 


10  ESTTEGEAL   CALCULUS.  [Art.  19. 

origin  as  a,  and  Ij-ing  on  the  opposite  side  of  the  origin ;  again 
repeat  the  operation, 

and  the  point  has  rotated  90°  further  ;  repeat  again, 

and  the  point  has  rotated  through  360°.  We  see,  then,  that  if 
the  subject  is  a  real  or  a  j^i'^re  imaginary  the  effect  of  performing 
on  it  the  operation  indicated  by  V  —  1  is  to  rotate  it  about  the 
origin  through  the  angle  90°.  We  shall  see  later  that  even  when 
the  subject  is  neither  a  real  nor  a  pure  imaginary,  the  effect  of 
operating  on  it  Anth  V  —  1  is  still  to  produce  the  rotation  just 
described. 

19.  The  sum,  the  product,  and  the  quotient  of  any  two  imagi- 
naries,  a  +  &  V  —  1  and  c  +  cZ  V  —  1 ,  are  imaginaries  of  the  typi- 
cal form. 

a  +  &V^n"  +  c  +  dV^     =a  +  c  +  (6  +  cOV^^.  [1] 

(a  +  W^)  (c  +  d  V^)  =ac-  bd  +  (be  +  ad)V^.         [2] 

a+b-\/^A  __  (a+h-\r~i)  (c— dV"^)  _ac+bd+(bc—ad)^/'^ 
c+d\/^l       (c+dV^)  (c-dV^)  c^  +  d' 

ac  -^bd      be  —  ad     , ■ 


c^  +  dr    '    c^  +  d^    ^~^-  ^^^ 

All  these  results  are  of  the  form  A  +  jBV— 1. 

20.    The    graphical    representation    we    have    suggested    for 
imaginaries  suggests  a  second  tj'pical  form  for  an  imaginary.  ■ 
Given  the  imaginary  cb  +  ?/V  — 1,  let  the  x>olar   coordinates  of 
the  point  P  which  represents  x-{-y'\/  —  1  be  r  and  (^. 

r  is  called  the  modidus  and  <j5)  the  argument  of  the  imaginary. 


Chap.  II.] 


IMAGINAEIES. 


11 


The  figure  enables  us   to   establish  veiy 
simple  relations  between  x,  y,  r,  and  cfi. 


x  =  rcos<j!> 
?/  =  r  sin  ^ 


;} 


d)=tan-i^. 

^  X 


x  +  y^—l  =  rcos<^+  (V  — l)rsin(^ 
=  r{coscJ3  +V  — l.sin^), 


[1] 
[2] 

[3] 


where  the  imaginary  is  expressed  in  terms  of  its  modulus  and 
argument. 

The  value  of  r  given  b}''  our  formulas  [2]  is  ambiguous  in 
sign  ;  and  ^  ma}-  have  any  one  of  an  iiifinite  number  of  values 
differing  by  multiples  of  tt.  In  practice  we  alwa3'S  take  the 
positive  value  of  r,  and  a  value  of  <^  which  will  bring  the  point 
in  question  into  the  right  quadrant.  In  the  case  of  any  given 
imaginar}^  then,  r  can  have  but  one  value,  while  4>  ma}'  have 
any  one  of  an  infinite  number  of  values  differing  by  2  tt. 


Examples. 

(1)  Find  the  modulus  and  argument  of  1 ;  of  V  — 1  ;  of  —  4  ; 
of  — 2V  — 1 ;  ofS  +  SV  — 1;  of2+4V— 1;  and  express  each  of 
these  quantities  in  the  form  ?'(coS(^ +V  — 1.  sin^). 

(2)  Show  that  every  positive  real  has  the  argument  zero ; 
every  negative  real  the  argument  tt  ;  every  positive  pure  imagi- 
nary the  ai'gument  ^ ;   and  every  negative  pure  imaginary  the 

argument  — . 

"^  2  .  .  • 

21.  If  we  add  two  imaginaries,  the  modulus  of  the  sum  is 
never  greater  than  the  sum  of  the  moduli  of  the  given  imagi- 
naries. 


12  INTEGRAL   CALCITLIJS.  [Art.  22. 

The  sum  of  a  +  5  V^  aud  c  +dV  — 1  is  a  +  c  +  (&  +  c?)  V  —  1. 
The  moduhis  of  this  sum  is  V(a  + c)' +  (&  +  ^0' ;  the  sum  of 
the  moduli  of  a  +6  V^  and  c  +d  V- 1  is  V  a^  +  &'  +  Vc^  +  d^. 
We  wish  to  show  that 

V(a  +  c)-  +  (6  +  d)'  -<  Va^+F+ V"?T^; 
tlie  sign  -<  meaning  "  equal  to  or  less  than." 
Now        '\/{a  +  cy  +  {b  +  dy  -<  ^/'aF+¥-\-  V c^  +  d,\ 


if  (a  +  c)2  +  (&  +  cZ)2  -<  cr  +  &'  +  2V(fr  +  &")  (c-  + d-)  +0^  +  cZ^, 


that  is,  if       ac  +  kZ  -<  Va^c^  +  a^d^  +  ft^c^  +  bhl' ; 

or,  squaring,  if 

a'  c'  +  2  abed  +  6^  d'  -<  a-  (?  +  a'  d-  +  &-  c^  +  ^  d' ; 

or,  if  0  -<  {ad  — be)-. 

This  last  result  is  necessarily  true,  as  no  real  can  have  a 
square  less  than  zero  ;  hence  our  proposition  is  established. 

22.  TJie  modidns  of  the  product  of  ttvo  imaginaries  is  the 
product  of  the  moduli  of  the  given  imaginaries.,  and  the  argument 
of  the  product  is  the  sum  of  the  arguments  of  the  imaginaries. 

Let  us  multiply 

?-i (cos  (^1  +  V  —  1 .  sin  <^i)     by     ?-2(cos  (^o  +  V  —  1 .  sin ^2)  ; 
we  get 
i\  >"2[cos  (jbi  cos  ^2— sia  <^i  sin  ^2+ V  —  l(sin  <^i  cos  (^2+ cos  <^i  sin  ^0)] , 

cos  <^i  cos  ^2  •—  sin  ^1  sin  ^2  =  cos  ( <^i  +  ^2)  ? 
sin  <^i  cos  ^2  +  cos  ^1  sin  ^2  =  sin  (</>!  +  ^2) 
hj  Trigonometrj' ;  hence 

ri  (cos  <^i  4-  V  —  1 .  sin  (^1)  r.,  (cos  (^1  +  V  —  1 .  sin  (^2) 
=  ri?'2  [cos  (<^i  4-  <^2)  +  V  - 1 .  sin  ((^1  +  (^o)  ] , 


Chap.  II.]  IMAGIjSTAKIES.  13 

and  our  result  is  in  the  tj-pical  form,  TiV^  being  tlie  modulus  and 
•^1  +  4>2  the  argument  of  the  product. 

If  each  factor  has  the  modulus  unity,  this  theorem  enables  us 
to  construct  very  easily  the  product  of  the  imaginaries  ;  it  also 
enables  us  to  show  that  the  interpretation  of  the  operation  V  —  1 , 
suggested  in  Art.  18,  is  perfectly  general. 

Let  us  operate  on  an}'  imaginary  subject, 

r(cos^4- V  — 1.  sin</)),  with  V— 1, 

that  is,  with  1  (  cos  ^  +  V  —  1 .  sin  ^  ] . 

The  modulus  r  will  be  unchanged,  the  argument  <^  will  be  in- 
creased by  -,  and  the  effect  will  be  to  cause  the  point  repre- 

senting  the  given  imaginary  to  rotate  about  the  origin  through 
an  angle  of  90°. 

23.  Since  division  is  the  inverse  of  multiplication, 
?-i(cos^i  + V  — 1.  sin</)i)  -i-  ?-2(cos(/)2  +  V— l.sint^a) 

will  be  equal  to 

-^  [cos  (c^i  -  <^2)  4-  V-1.  sin(<^i  —  <^2)], 

since  if  we  multiply  this  b}'  ?-2(cos^2+  V— 1.  sin  ^0)5  according 
to  the  method  established  in  Art.  22,  we  must  get 

rj  (cos  </>!  +  V  —  1 .  sin  ^^ . 

To  divide  one  imaginary  by  another,  we  have  then  to  take  the 
quotient  obtained  by  dividing  the  modulus  of  the  first  by  the 
modulus  of  the  second  as  our  required  modulus,  and  the  argu- 
ment of  the  first  niinus  the  argument  of  the  second  as  our  new 
argument. 

24.  If  w.e  are  dealing  with  the  product  of  n  equal  factors,  or, 
in  other  words,  if  we  are  raising  r(cos<^  +  V— l.sin^)  to  the 


14  LNTEGEAIi   CALCULUS.  [Art.  25. 

nth  power,  n  being  a  positive  whole  number,  we  shall  have,  by 
Art.  22, 

[r(cos  ^  +  V  —  1 .  sin  ^)  ]"  =  r"(cos?i  <j>  +  V  — 1.  sin?i  cji) .       [1] 

If  r  is  unit^',  we  have  merel}^  to  multiply  the  argument  by  n, 
without  changing  the  modulus  ;  so  that  in  this  case  increasing 
the  exponent  by  unit}'  amounts  to  rotating  the  point  represent- 
ing the  imaginary  through  an  angle  equal  to  ^  without  changing 
its  distance  from  the  origin. 


25.    Since  extracting  a  root  is  the  inverse  of  raising  to  a 
power, 

V[r(cos<^  +  V^.sin<^)]  =  Trf  cos^  + V^.sin^j  ;       [1] 

for,  by  Art.  24, 

-^•(eos-+V"^.sin-) 
^    \       n  '  nj 


—  ?'(coS(^  + V  — 1.  sin<^). 


Example. 

Show  that  Art.  24  [1]  holds  even  when  n  is  negative  or 
fractional. 

26.  As  the  modulus  of  every  quantity,  positive,  negative, 
real,  or  imaginar}',  is  positive,  it  is  alwa^'s  possible  to  find  the 
modulus  of  any  required  root ;  and  as  this  modulus  must  be  real 
and  positive,  it  ca7i  never,  in  an}-  given  example,  have  more  than 
one  value.  We  know  from  algebra,  however,  that  ever}'  equa- 
tidn  of  tlie  nth  degree  containing  one  unlinown  has  n  roots,  and 
that  consequently  ever}'  number  must  have  n  nth  roots.  Our 
formula.  Art.  25  [1],  appears  to  give  us  but  one  nth  root  for 
any  given  quantity.     It  must  then  be  incomplete. 

We  have  seen  (Art.  20)  that  while  tlie  modulus  of  a  given 
imaginary  has  but  one  value,  its  argument  is  indeterminate  and 
may  have  any  one  of  an  infinite  number  of  values  whicli  differ  b}' 
multiples  of  27r.     If  <^o  is  one  of  these  values,  the  full  form  of 


Chap.  II.]  IMAGINAEIES.  15 

the  imaginary  is  not  ?'(cos  4>o  +  V  —  1.  sin<^o)  5  ^.s  we  have  hitherto 
written  it,  but  is 

r  [cos (^0  +  2  mir)  -|-  V — 1 .  sin (<^o  +  2  mir) ] , 

where  m  is  zero  or  any  whole  number  positive  or  negative. 
Since  angles  differing  by  multiples  of  2  tt  have  the  same  trigo- 
nometric functions,  it  is  easily  seen  that  the  introduction  of  the 
term  2  mir  into  the  argument  of  an  imaginar}^  will  not  modif}' 
any  of  our  results  except  that  of  Art.  25,  which  becomes 


Vr  [eos(<^o+  2?ft7r)  +  V— 1-  sin((^o+  2?)i7r)] 


=  ^/- 


Mo  2  7r\  , .     Mo  27rY 

COS \-m  —    +  v  —  1 .  sm \-m  — 


[1] 


Giving  VI  the  values  0,  1,  2,  3  ,  n  —  1,  w,  n  -\-l,  success- 
ively", we  get 


^0        2  7r      (j>Q          2tv      4>o           2Tr  <^o    , 

5 1 5 1- ^ — 7 ho —  5    — + 


n—1  2- 


n      n        n       n 

^+2.,    ^  +  -  +  2., 
n  n        n 

as  arguments  of  our  7ith  root. 

Of  these  values  the  first  n,  that  is,  all  except  the  last  two, 
correspond  to  different  points,  and  therefore  to  different  roots  ; 
the  next  to  the  last  gives  the  same  point  as  the  first,  and  the 
last  the  same  point  as  the  second,  and  it  is  easil}'  seen  that  if  we 
go  on  increasing  m  we  shall  get  no  new  points.  The  same  thing 
is  true  of  negative  values  of  m. 

Hence  we  see  that  every  quantity^  real  or  imaginary ^  has  n 
distinct  nth  roots,  all  having  the  same  modulus,  but  with  argu- 
ments diflfering  by  multiples  of  ^^ . 

27.  Anj^  positive  real  differs  from  unity  only  by  its  modulus, 
and  any  negative  real  differs  from  —1  onl}'  by  its  modulus.  All 
the  ?ith  roots  of  any  number  or  of  its  negative  may  be  obtained 


16 


INTEGRAL   CALCULUS. 


[Art.  27. 


hj  multiplj'ing  the  nth  roots  of  1  or  of  —  1  by  the  real  positive 
wth  roots  of  the  number. 

Let  us  consider  some  of  the  roots  of  1  and  of  —  1 ;  for  ex- 
ample, the  cube  roots  of  1  and  of  —1.  The  modulus  of  1 
is  1    and  its  aro-ument  is  0.    The  modulus  of  each  of  the  cube 

roots  of  1  is  1,  and  their  arguments  are  0,  — ,  and  —  ;  that  is, 
0°,  120°,  and  240°.  The  roots  in  question,  then,  are  repre- 
sented by  the  points  Pj,  Pa,  ^s,  in  the  figure.     Their  values  are 

l(cosO  + V^.  sinO), 
1  (cos  120°  +  V^.  sin  120°), 
and  I(cos240°  +  V^.sin240°), 
or  1,  -i+.^V^l,   -i-^V^. 

The  modulus  of  —1  is  1,  and  its 

argument  is  tt.     The  modulus  of  the 

cube  roots  of  —1  is  1,   and  their  arguments  are  ^,  J  +  -^, 

!r  +  i5,  that  is,  60°,  180°,  300°.  The  roots  in  question,  then, 
3       3 

are  represented  by  the  points  Pi,  Pg, 

Pg,  in  the  figure.     Their  values  are 

-^-\-    2      V  — 1,     — 1,    -2  —    2     V — 1. 

Examples. 

(1)  What  are  the  square  roots  of 
1  and  —  1  ?  the  4th  roots  ?  the  5th 
roots  ?   the  6th  roots  ? 


(2)  Find  the  cube  roots  of  —  8  ;  the  5th  roots  of  32. 


(3)  Show  that  an  imaginary'  can  have  no  real  ?ith  root ;  that 
a  positive  real  has  two  real  ?ith  roots  if  n  is  even,  one  if  n  is 
odd ;  that  a  negative  real  has  one  real  «th  root  if  n  is  odd,  none 
if  n  is  even. 


Chap.  II.]  IMAGINAEIES.  17 

28.  Imaginaries  having  equal  moduli,  and  arguments  differing 
only  in  sign,  are  called  conjugate  imaginaries. 

r(cos<^  + V  — l.siu^),  and  r[cos(  — </>)  +  V  — l.sin(  — ^)], 
or  r(cos  (^  —  V—  1 .  sin  (^)  are  conjugate. 

They  can  be  written  x  +  y  V  — 1  and  x  —  7j\/—l,  and  we  see 
that  the  points  corresponding  to  them  have  the  same  abscissa, 
and  ordinates  which  are  equal  with  opposite  signs. 

ExAjMPLES. 

(1)  Prove  that  conjugate  imaginaries  have  a  real  sum  and  a 
real  product. 

(2)  Prove,  by  considering  in  detail  the  substitution  of 
a  +  6  V— 1  and  a  —  &V— 1  in  turn  for  x  in  an}^  algebraic  poly- 
nomial in  X  with  real  coefficients,  that  if  any  algebraic  equation 
with  real  coefficients  has  an  imaginary  root  the  conjugate  of  that 
root  is  also  a  root  of  the  equation. 

(3)  Prove  that  if  in  an}^  fraction  where  the  numerator  and 
denominator  are  rational  algebraic  polynomials  in  x,  we  substi- 
tute a  +  6V— 1  and  a  — &V  — 1  in  turn  for  x,  the  results  are 
conjugate. 

Transcendental  Functions  of  Imaginaries. 

29.  We  have  adopted  a  definition  of  an  imaginary  and  laid 
down  rules  to  govern  its  use,  that  enable  us  to  deal  with  it,  in 
all  expressions  involving  only  algebraic  operations,  precisely  as 
if  it  were  a  quantity.  If  we  are  going  further,  and  are  to  sub- 
ject it  to  transcendental  operations,  we  must  carefully  define 
each  function  that  we  are  going  to  use,  and  establish  the  rules 
which  the  function  must  obey. 

The  principal  transcendental  functions  are  e^,  logic,  and  sinx, 
and  we  wish  to  define  and  stud}'  these  when  x  is  replaced  by  an 
imaginary  variable  z. 

As  our  conception  and  treatment  of  imaginaries  have  been 
entirely  algebraic,  we  naturall}'  wish  to  define  our  transcendental 


18  INTEGRAL  CALCULUS.  [Art.  30. 

functions  b}'  the  aid  of  algebraic  functions  ;  and  since  we  know 
that  the  transcendental  functions  of  a  real  variable  can  be  ex- 
pressed in  terms  of  algebraic  functions  onlj'  by  the  aid  of  infinite 
series,  we  are  led  to  use  such  series  in  defining  transcendental 
functions  of  an  imaginary  variable  ;  but  we  must  first  establish 
a  proposition  concerning  the  convergency  of  a  series  containing 
imaginar}'  terms. 

30.  If  the  moduli  of  the  terms  of  a  series  containing  imaginary 
terms  form  a  convergent  series,  the  given  series  is  convergent. 

Let  Uq -{- III -\- ih -^ +  w»  + be  a  series  containing  imagi- 

nar}'  terms. 

Let  

^(o  =  i2o(cos*o+  V  — 1.  sincEJo),  ti^  =  i?i  (cos  $i + V  —  1 .  sin  ^j),  &c., 

and  suppose  that  the   series  i?o  +  -Ki  +-R2  + +  -S«  + is 

convergent ;  then  will  the  series  Uq-}-  r<i+  t<2+ be  convergent. 

The  series  jRo  +  -Ki+ is  a  convergent  series  composed  of 

positive  terms  ;  if  then  we  break  up  the  series  into  parts  in  any 
way,  each  part  will  have  a  definite  sum  or  will  approach  a  defi- 
nite limit  as  the  number  of  terms  considered  is  increased  in- 
definitely. 

The  series  ^(o  +  Ui-{-ti2  + Un-i- can  be  broken  up  into 

the  two  series 

jRoCos$o  +  -?2iCos$i  +  i22Cos$2  4- +  -^«cos$„  + (1) 

and 

V^(i?osin$o  +  -?^iSin<E>i+i?2sin$2-l •  +  i2„sin$„-f ).  (2) 

(1)  can  be  separated  into  two  parts,  the  first  made  up  only 
of  positive  terms,  tli€  second  only  of  negative  terms,  and  can 
therefore  be  regarded  as  the  difference  between  two  series,  each 
consisting  of  positive  terms.     Each  term  in  either  series  will  be 

a  term  of  the  modulus  series  ^0+^1  +  ^2  + multiplied  by 

a  quantit}'  less  than  one,  and  the  sum  of  n  terms  of  each  series 
will  therefore  approach  a  definite  limit,  as  n  increases  indefi- 
nitely. The  series  (1),  then,  which  is  the  abscissa  of  the  point 
representing  the  given  imaginary  series,  has  a  finite  sum. 


Chap.  II.]  IMAGINAEIES.  19 

In  the  same  way  it  ma}'-  be  shown  that  the  coefficient  of  V— 1 
in  (2)  has  a  finite  sum,  and  this  is  the  ordinate  of  the  point 
representing  the  given  series.  The  sum  of  n  terms  of  the  given 
series,  tlien,  approaches  a  definite  limit  as  n  is  increased  indefi- 
nitely, and  the  series  is  convergent. 

31.    We  have  seen  (I.  Art.  133  [2])  that 

e^  =  l+^-f-^  +  ^-)-^+ [1] 

12!3!4! 

when  X  is  real,  and  that  this  series  is  convergent  for  all  values  of  x. 
Let  us  define  e%  where  z  =  x-{-y\/  —  l,  by  the  series 

12!      3!      4!  ■-  -^ 

This  series  is  convergent,  for  if  z  =  r(cos  ^  +  V—  1 .  sin <^)  the 
series 

i  +  !:+i!+il+i!.+ 

1      2!      3!      4! 

made  up  of  the  moduli  of  the  terms  of  [2]  is  convergent  hj 
I.  Art.  133,  and  therefore  the  value  we  have  chosen  for  e"  is  a 
determinate  finite  one. 

Write  a;  +  ?/V— 1  for  2,  and  we  get 

c'+v^/=i^l  I  a^+yV-l  ^  (a;+yV3i)2  ^  (a;_^yV~i)3  ^  

The  terms  of  this  series  can  be  expanded  by  the  Binomial 
Theorem.  Consider  all  the  resulting  terms  containing  any  given 
power  of  X,  say  x^ ;  we  have 

pr^    1    ^     2!     ^     3!     ^      ^     ;:!     ^""r 

or,  separating  the  real  terms  and  the  imaginary  terms, 

^(i_i!  +  ^_/  + >, 

2^1  2!4!      6!  ^ 

pi  ^^      3!      5!      7!  ^' 


20  INTEGRAL   CALCULUS.  [Art.  32. 

or  —  (cosy  +  V^.  sin?/),  b}'  I.  Art.  134. 

p  ! 

Giving  p  all  values  from  1  to  co  we  get 

gx+y./^  =  (cos  ?/  +  V^.  siny)  (i+Y  +  fr  +  fr  +  fT'^ ^ 

=  e^  (cosy  +  V—1.  sin?/),  [4] 

which,  b}'  the  way,  is  in  one  of  our  tj-pical  imaginary  forms. 

If  a;=0,  in  [4], 
we  get  e^^-i  =  cos?/  + V^.siny, 

which  suggests  a  new  way  of  writing  our  typical  imaginary; 

namely,      *  

r  (cos  ^  +  V  —  1 .  sin  ^)  =  re*^^-\ 

32.    We  have  seen  that 

let  us  see  if  all  imaginary  powers  of  e  obey  the  laiv  of  indices; 

that  is,  if  the  equation 

g«g«^g«  +  «  ["ij 

is  universally  true. 

Let  ?(  =  a;i  +  ?/iV— 1    and    'y  =  3^2  +  2/2^—1, 

then  e"=  e^i  +  2/1  -^^  =  e«i (cos?/i  +  V—  1  •  sin?/i) , 

6"=  ea:o  +  2/oN/^=  ea^^^-cosya  +  V— 1.  siuyg)  ? 

g«  g«  _  ga;i  ga;^  ["cog  (^/^  -f  3/2)  +  V  —  1  •  sin  (?/i  +  y,)  ] 

=  e^i+ ^'2  [cos  (yi  +  Vi)  +  V^ .  sin  (yi  +  2/2)  ] 

_  ga;i  +  a;2  +  (2/1  +  2/2)  V^ 

__    gM  +  » 

and  the  fundamental  property  of  exponential  functions  holds  for 

imaginaries  as  well  as  for  reals.  > 

Example. 
Prove  that  a'^a"  =  ^4"  +  "  when  u  and  v  are  imaginary. 


Chap.  II.]  IMAGIISTAKIES.  21 

LogaritJimic  Functions. 

33.  As  a  logarithm  is  the  inverse  of  an  exponential,  we  ought 
to  be  able  to  obtain  the  logarithm  of  an  imaginar}-  from  the 
formula  for  e'=+^"'^^    We  see  readily  that 

z  =  r  (cos<^  +  V^.  sin<^)  =  eiogr+0^ 
whence  logz  =  logr  +  ^  V  —  1 ; 

or,  more  strictly,  since 

z  =  r[cos  (<^o  +  2n7r)  +  V  — l.sin(</>o+  2n7r)], 

•log2  =  logr+(<Ao  +  2n7r)  V"=^  [1] 

where  n  is  an}^  integer. 

If  z  =  x  +  y  V— 1,  r  =  Va^nTp,  and  ^  =  tan~^^ ; 

X 

whence  logz  =  *log  (x^  +  ^/S)  +  V^.tan-^^.  [2] 

X 

Each  of  the  expressions  for  log  z  is  indeterminate,  and  repre- 
sents an  infinite  number  of  values,  differing  by  multiples  of 
27rV'^. 

This  indeterminateness  in  the  logarithm  might  have  been  ex- 
pected a  x>riori,  for 

g27rv/^i^gQg27r+V^.sin27r=  1,       by  Art.  31. 

Hence,  adding  2  ir  V—  1  to  the  logarithm  of  an}^  quantity  will 
have  the  effect  of  multiptying  the  quantity  by  1 ,  and  therefore 
win  not  change  its  value.  , 


Example. 

Show  that  if  an  expression  is  imaginary,  all  its  logarithms  are 
imaginary ;  if  it  is  real  and  positive,  one  logarithm  is  real  and 
the  rest  imaginary  ;  if  it  is  real  and  negative,  all  are  imaginary. 


22  INTEGRAL  CALCULUS.  [Akt.  34. 

Trigonometric  Functions. 

34.  If  z  is  real, 

sin.  =  . --  +  --_  + [1] 

eos.  =  l-|l  +  |l-^+ [2] 

byl.  Art.134.  ^-      ^-      '^ ' 

If  2  =  r(cos^+ V— l.sinc^), 

the  series  of  the  moduli, 

n\0  ntO  rt»7 

^"  +  FT  +  ^  +  ^+ , 

3  !      0  !      7  ! 

1  +  if + j1  4.  i!  _f- 

2 !      4!      6!  ' 

are  easily  seen  to  be  convergent ;  therefore  if  z  is  imaginary,  the 
series  [1]  and  [2]  are  convergent.  We  shall  take  them  as  defi- 
nitions of  the  sine  and  cosine  of  an  imaginary. 

Example. 

From  the  formulas  of  Art.  31,  and  from  Art.  34  [1]  and  [2], 
show  that 

gzV^  _  COS2  _|_  V  — 1.  sin2, 

and  e'^"^"^  =  coss;  —  V  — 1.  sin^,     for  all  values  of  z. 

35.  From  the  relations 

g^v-i _  gQg J, _|_  y_][^ sins;, 

we  get  C0S2  = — ,  [1] 

sm2  = — ,  12  I 

for  all  values  of  0. 


Chap.  II.] 
Let 

cos(cc+?/V— 1)  = 


IMAGINAEIES. 

z  —  x  +  y'\l —1. 


g^v/-l_y^g-x^^l  +  2 


23 


_(cosx+V— l.siaa;)e~^+(cosa;— V— l.sina;)e^ 

~  2  ' 

by  Art.  34,  Ex., 


=  COS  a;  — ■ —  V  —  1 .  siiiic 

2 

In  the  same  wa}^  it  ma}^  be  shown  that 


[3] 


.  ,   I — 7^   (cosa;+ V  — 1.  sina;)e"^—  (cosa;  —  V— l.sina;)e^ 
l(.x+?/V-1)=^ ~ '- — ^ — 

2V-I 


+  V-1. 


cos  X  ■ 


[4] 


If  z  is  real  in  [1]  and  [2] ,  we  have 


cos  X  = 


+  e- 


2 


„xV-l     „-lV^l 


V^. 


lfz  =  y  V  — 1,  and  is  a  pure  imaginary, 


cosy 


1  = 


2 


sin  ?/  V  —  1  = 


2 


V^; 


[5] 
[6] 


whence  we  see  that  the  cosine  of  a  pure  imaginary  is  real,  while 
its  sine  is  imaginary. 

By  the  aid  of  [5]  and  [6] ,  [3]  and  [4]  can  be  written : 

cos  (x-\-yy/  —  1)  =  cosoJcosyV— 1  —  sinxsin?/ V— 1,     [7] 
sin  (x  +  2/ V— 1)  =  siniccos?/ V— 1  +  cosccsin^/V  — 1.     [8] 


24  INTEGEAL  CALCULUS.  [Art.  36. 

Examples. 

(1)  From  [1]  and  [2]  show  that  sin^2:+ cos^2;=  1. 

(2)  Prove  that 

cos  (?i  -{-v)=  COS  M  cos 'y  —  shaitsmv, 
sin  (u  -\-v)  =  sinttcosi;  +  cosMsmi>, 
where  u  and  v  are  imaginary. 

The  relations  to  be  proved  in  examples  (1)  and  (2)  are  the 
fundamental  formulas  of  Trigonometry',  and  they  enable  us  to 
use  trigonometric  functions  of  imaginaries  precisely  as  we  use 
trigonometric  functions  of  reals. 

Differentiation  of  Functions  of  Imaginaries. 

36.  A  function  of  an  imaginary  variable, 

is,  strictly  speaking,  a  function  of  two  independent  variables, 
X  and  y  ;  for  we  can  change  z  by  changing  either  x  or  ?/,  or  both 
X  and  y.  Its  differential  will  usually  contain  dx  and  dy,  and  not 
necessarily  dz ;  and  if  we  divide  its  differential  by  dz  to  get  its 

dy 
derivative  with  respect  to  2,  the  result  will  generallj-  contain  —■, 

OjOC 

which  will  be  wholl}'^  indeterminate,  since  a;  and  y  are  entirely 
independent  in  the  expression  x  +  y  V  — 1.  It  may  happen, 
however,  in  the  case  of  some  simple  functions,  that  dz  will  appear 
as  a  factor  in  the  differential  of  the  function,  which  in  that  case 
will  have  a  single  derivative. 

37.  In  differentiating^  the  V  — 1  may  be  treated  like  a  con- 
stant; for  the  operation  of  finding  the  differential  of  a  function 
is  an  algebraic  operation,  and  in  all  algebraic  operations  V  — 1 
obej^s  the  same  laws  as  an^^  constant. 


CH-iP.  II.] 


EMAGIiSrAEIES. 


25 


Example. 

Prove  that  d(xV^)  =  2x V^ .  dx ; 

and  that  dV— 1.  sma;=  'sj  —  l.cosx.dx 

We  have,  by  the  aid  of  this  principle, 
if  z  =  x-\-y\r—i, 

dz=  dcc  +  V  — 1.  dy;  [1] 

if       z  =  r  (cos<^  +  V  — 1.  sine/)), 

dz  =  d9'(cos </)  +  V  —  1 .  sin (^)  +  rdcji (sin  <^  +  V  — 1 .  cos  ^) 
=  (dr  +  ?'  V^ .  d<f>)  (cos  <^  +  V^ .  sin  cjy).  [2] 

38.    Let  us  now  consider  the  differentiation  of  2"',  e^,  logs, 
sins,  and  cos 2;. 

Let  z  =  r  (cos  <^  +  V— 1 .  sin  <^) , 

then 
gm  _  ,''»(cos?n^  +  V^.  sinm^) ,  hj  Axt.  24  [1]  ; 

dz""  =  jjir™"^  dr  (cos  77i<^  +  V  —  1 .  sin  ??i</))  +  onr""  d(f>  (  —  sin  m  <^ 
+  V  — l.cosmc/)), 

d^'"  =  7/ir™-^  [cos  (7)1—1)  </)  +  V  — l.sin(7?i  — 1)  c/>]  (cosej!) 
+  V  — 1.  sin  (j>)dr 

+  m?''"[cos  (7?i— 1)  <^  +  V  — l.sin(m  — 1)0]  (coscfi 
+ V^.  sin<^)  ^f^.dcji, 

dz^  =  mr""'^  [cos  (m  —  1 )  c/)  -f-  V  —  1 .  sin  (m — 1 )  (^]  (dr 

_j-  ?-V  — 1  .dcf>)  (cos</>  +  V— 1.  sin<^), 
dz"' =  mz'^-hlz,  [1]     byArt.  37[2], 

^  =  mz"'-\  [2] 

and  a  power  of  an  imaginarj^  variable  has  a  single  derivative. 

Example. 
Show  that  [1]  and  [2]  hold  for  all  powers  of  z. 


26  ESTTEGEAL   CALCULUS.  [Art.  39. 

39.    If     2=a;  +  ?/V^, 

e'^  =  e'^(cos2/+V  — l.sin?/),  by  Art.  31  [1], 

de'  =  e'dx ( cos ?/  -f  V^ .  svay) -\- e" {—  sin y 
-\-'\/—l.cosy)dy, 

de^  =  e==(cos2/+V  — 1.  sin?/)  (dx+V— 1.  dy), 
de^=e''dz,  [1] 

^^'  =  e^  '  [2] 


d^ 


Show  that 


Example. 

da'^  =  a^  los:  a.  dz. 


40.    If  «  =  ^(cos<^  + V-l.sin(^), 

log2:  =  log?'  +  ^  V  — 1,  by  Art.  33, 

d\ogz  = 1-  V  —  1 .  d<^  = ■ ^, 


g^l^^^ ^  (r7r+rV-l.r?.^)(cosj>+V-l.sin<^) 
r(cos(^  +  V  — 1.  sin^) 


d  loaf  2  = 


dz 


dlogz  _  1 
dz         z 


[1] 
[2] 


41.      sins:: 


dsins;  = 


2V-1 


by  Art.  35  [2], 


^.^i_^^_.^i 


2  V-1 


g^«Ari_|_g_^V_i 


V-1.CZ2 


dz, 


dsin2  =  cosz.dz. 


by  Art.  35  [1], 
[1] 


Chap.  II.]  niAGINAEIES.  27 


cosz  = , 


dcos«  =  ^ ^ V-l.cfe  = —=z cfe, 

2  2  V^ 

dcosz  =  —  sin2!.cfe.  [2] 


42.  "We  see,  then,  that  we  get  the  same  formulas  for  the  dif- 
ferentiation of  simple  functions  of  imaginaries  as  for  the  dif- 
ferentiation of  the  corresponding  functions  of  reals.  It  follows 
that  our  formulas  for  du'ect  integration  (I.  Art.  74)  hold  when  x 
is  imaginar}^. 

Hyperbolic  Functions. 

43.  We  have  (Art.  35  [5]  and  [6]) 


cos  a; 


V^  =  ,^'  +  ^"' 


2       ' 

and  sina?  V^  ^  e"  -  ^"V^, 

2 

where  a;  is  real.      — r —  is  called  the  hj^perbolic  cosine  of  x, 

and  is  written  Ch  x  ;   and  — - —  is  called  the  hj^perbolic  sine 

id 

of  x^  and  is  written  Sha; : 

Sha;  =.  ^^  ~  ^~'  =  -  V^.  sina;  V"^,  [1] 

Cha;  =  ^'  +  ^~'  =  cos  a;  V^.  [2] 

The  hj^perbolic  tangent  is  defined  as  the  ratio  of  Sh  to  Ch ; 
and  the  hjiDerbolic  cotangent,  secant,  and  cosecant  are  the  re- 
ciprocals of  the  Th,  Ch,  and  Sh  respectively. 

These  functions,  which  are  real  when  x  is  real,  resemble  in 
their  properties  the  ordiuar}^  trigonometric  functions. 


28  INTEGRAL    CALCULUS. 

44.   For  example, 


for 
and 


Ch2a:-Sh2a;=l; 

gSx  _|_  2  ^  Q-2Z 


Ch^x 


Sh2£c  = 


e2x_  2  _|_  e- 


[Art.  44. 


[1] 


Examples. 

(1)  Prove  that  1 -Tli^'c=  Sch^'c. 

(2)  Prove  that  1  -  Cth2aj=  Csch^o;. 

(3)  Prove  that  Sh(iB  +  ?/)=  ShicCh2/  + Cha^Sh?/. 

(4)  Prove  that  Ch(a;  + ?/)  =  Ch.'cCh?/  + ShxShy. 


45. 


dSha;  =  d- 


e'  -\-e' 


dx. 


dShx  =  Chx.dx. 


Examples. 

(1)  Prove  dChx  =  Shx.dx. 

dThx=  Sch^x.dx. 
dCtha;=  —Csch^x.dx. 
,.  dSehcc=  —SehxThx.dx. 
dCschx  =  —  Cscha;Ctha.\da;. 

46.    We  can  deal  with  anti-h}^erbolic  functions  just  as  with 
anti-trigonometric  functions. 
To  find  dSh-^a;. 


Let 
then 


u  —  Sh'^o;, 
x=  Sh^t, 
dx—  Chw.cZw, 


Chap,  n.]  LMAGESTAEIES.  29 

du= , 

Chu 


Chu=  Vl  +  Sh^w,  by  Art.  4  [1] , 

Examples. 
dCh-^x  = 
dTh-^x  = 


Prove  the  formulas 

dx 


\     CI  Sell  ''^x  =  — 
dCsch.~^x=— 


^/x''-l 
dx 

dx 


x  Vl— 0^ 

dx 
X  Var  + 1 


47.   The  anti-hj-perbolic  functions   are  easily  expressed  as 
logarithms. 

Let  ic=Sh~'^x, 


then  a;  =  Sh  tt  = 


e"  —  e' 


2x  =  e" , 

e"-x=  ±  VlTa?, 

e"  =  «  ±  VIT^ ; 


30  ESTTEGBAL   CALCULUS.  [Akt.  48. 

as  e"  is  necessarily  positive,  we  may  reject  the  negative  value  in 
the  second  member  as  impossible,  and  we  have 

u  =  log(.i;  +  Vl  +  it--) , 
or  Sh-iic  =  log(x4- Vr+^).  [1] 

EXAJEPLES. 

Prove  the  formulas 

Ch-^a;  =  log(.T  +  Va;-  —  1) . 


Th-^aj  =  ilog 


l+x 


1  —  X 


Csch-'a;=Iog('l  +  Ji+A 


48.  The  principal  advantage  ai'isiug  from  the  use  of  h}^er- 
bolic  functions  is  that  they  bring  to  light  some  curious  analogies 
between  the  integrals  of  certain  irrational  functions. 

From  I.  Art.  71  we  obtain  the  formulas  for  direct  integration. 


s 


dx  .    _i  rt-t 

=  sm  ^x.  [1] 


V 1  -  ar 
'^"^         =tan-ia;.  [2] 


1  +  it^ 
dx 


X  Va;^—  1 

From  Art.  46  we  obtain  the  allied  formulas 
dx 


sec"^a;.  [3] 


Vl+a.-^ 
dx 


=  Sh-^x  =  log{x  +  VI+^).  W 

=  Ch-ia;  =  log(a;  +  V^^^^).  E^] 


Chap.  11.] 

EVIAGINAEIES. 

r  dx 

J  l-x^ 

r    dx 

-=Sch-ia)   =looYi4-^ 

x^l—x^ 


31 

[6] 

[7] 


-  f^^  =  Csch-^^  =  log  fi  +  aE+iY  [8] 


32  INTEGRAL    CALCULUS.  [Art.  49. 


CHAPTER    III. 

GENEKAL  METHODS   OF  INTEGEATING. 

49.  We  have  defined  the  integral  of  any  function  of  a  single 
variable  as  the  function  which  has  the  given  function  for  its 
derivative  (I.  Art.  53)  ;  we  have  defined  a  definite  integral  as 
the  limit  of  the  sum  of  a  set  of  differentials ;  and  we  have  shown 
that  a  definite  integral  is  the  difference  betioeen  two  values  of  an 
ordinary  integrcd  (I.  Art.  183) . 

Now  that  we  have  adopted  the  diflTerential  notation  in  place  of 
the  derivative  notation,  it  is  better  to  regard  an  integral  as  the 
inverse  of  a  differential  instead  of  as  the  inverse  of  a  derivative. 
Hence  the  integral  of  fx.dx  will  be  the  function  whose  differ- 
ential is  fx.dx;  and  we  shall  indicate  it  by  i  fx.dx.  In  our  old 
notation  we  should  have  indicated  precisely  the  same  function  by 
I  fx ;  for  if  the  derivative  of  a  function  is  fx  we  know  that  its 
differential  is  fx.dx. 

50.  If /c  is  an}'  function  whatever  of  a;,  fx.dx  has  an  integral. 
For  if  we  construct  the  curve  whose  equation  is  y=fx,  we  know 
that  the  area  included  by  the  curve,  the  axis  of  X,  any  fixed 
ordinate,  and  the  ordinate  corresponding  to  the  variable  x,  has 
for  its  difl'erential  ydx.,  or,  in  other  words,  fx.dx  (I.  Art.  51). 
Such  an  area  always  exists,  and  it  is  a  determinate  function  of  x, 
except  that,  as  the  position  of  the  initial  ordinate  is  wholl}'  arbi- 
trary, the  expression  for  the  area  will  contain  an  arbitrary  con- 
stant.   Thus,  if  Fx  is  the  area  in  question  for  some  one  position 

.of  the  initial  ordinate,  we  shall  have 

Cfx.dx  =  Fx-\-G, 
where  C  is  an  arbitrary  constant. 


Chap.  III.]  GENERAL  METHODS   OF  INTEGEATING.  33 

Moreover,  Fa;  +  C  is  a  complete  expression  for  j  fx.dx  ;  for  if 
two  functions  of  x  have  the  same  differential,  they  have  the  same 
derivative  with  respect  to  x,  and  therefore  they  change  at  the 
same  rate  when  x  changes  (I.  Art.  38)  ;  they  can  differ,  then, 
at  any  instant  only  by  the  difference  between  their  initial  values, 
which  is  some  constant. 

Hence  we  see  that  every  expression  of  the  form  fx.dx  has  an 
integral^  and,  except  for  the  presence  of  an  arbitrary  constant, 
but  one  integral. 

51.  We  have  shown  in  I.  Art.  183  that  a  definite  integral 
is  the  difference  between  two  values  of  an  ordinaiy  integral,  and 
therefore  contains  no  constant.  Thus,  if  Fx-\-G  is  the  integral 
of  fx.dx, 

Jfx.dx  =  Fb  —  Fa. 
a 

In  the  same  way  we  shall  have 

Cfz.dz  =  Fb-Fa; 

and  we  see  that  a  definite  integral  is  a  function  of  the  values 
between  tohich  the  sum  is  taken  and  not  of  the  variable  with 
respect  to  which  we  integrate. 

Since 


ffx.dx^Fa  —  Fb, 
I  fx.dx  =  —  I  fx.dx. 

h  Ja 


Example. 


fx.dx  +  I  fx.dx  =  I  fx.dx. 


52.  In  what  we  have  said  concerning  definite  integrals  we 
have  tacitly  assumed  that  the  integral  is  a  continuous  function 
between  the  values  between  which  the  sum  in  question  is  taken. 
If  it  is  not,  we  cannot  regard  the  whole  increment  of  Fx  as  equal 


34 


INTEGRAL    CALCULUS. 


[Art.  53. 


r 
to  the  limit  of  the  sum  of  the  partial  infinitesimal  increments, 
and  the  reasoning  of  I.  Art.  183  ceases  to  be  valid. 
Take,  for  example,    )     — -. 


and  apparently 


j-=j.-c?.  =  _  =  --,     by  I.  Art.  55  (7); 


But  j     — ;-  ought  to  be  the  area  between  the  curve  ?/  =  — ,  the 
axis  of  X,  and  the  ordinates  corresponding  to  cc  =  1  and  ic  =  —  1 , 

which  evidently  is  not  —  2  ;  and  we 

see  that  the  function  —  is  discon- 

x^ 
tinuous  between  the  values  x=  —I 
and  x  =  \. 

The  area  in  question  which  the 
definite  integral  should  represent  is 
.J-       easil}'  seen  to  be  infinite,  for 


-1      0 


J-\    ar      e  J e   x^      €. 


and  each  of  these  expressions  increases  without  limit  as  e  ap- 
proaches zero, 

53.  Since  a  definite  integral  is  the  difference  between  two 
values  of  an  indefinite  integral,  what  we  have  to  find  first  in  any 
problem  is  the  indefinite  integral.  This  ma}'  be  found  by  in- 
spection if  the  function  to  be  integrated  comes  under  any  of  the 
forms  we  have  already  obtained  b}'  differentiation,  and  we  are 
then  said  to  integrate  directl}".  Direct  integration  has  been  illus- 
trated, and  the  most  important  of  the  forms  which  can  be  in- 
tegTated  directly  have  been  given  in  I.  Chapter  V.  For  the  sake 
of  convenience  we  rewrite  these  forms,  using  the  diflferential 
notation,  and  adding  one  or  two  new  forms  from  our  sections  on 
hyperbolic  functions. 


Chap.  III.]     GENERAL  METHODS   OE  INTEGEATIKG.  35 

/^,n  + 1 
x''dx  =  - . 
71  +  1 

/dx     , 
—  =  logic. 
X 

/a'dx=z-^-. 
loga 

I  e'dx^  e". 

I  smx.dx=  —cos a;. 

I  cosx.dx  =  sin  a;. 

I  tana.\cZa;=  —  logcosa;. 

I  ctua;.da;=  logsino;. 

I  =  sm  ^a;. 

c/  Vl-ic^ 

I     ,  =  Sh-^a;  =  log(a;  +  Vl  +  x^) . 

*^  Vl  +  a^ 

/-.  =  Qh.-'^x  =  log  (a;  +  V.«^  —  1) . 

Va.-^  —  1 

/r^-Th-..=iiogi±|. 
/- 


da;  , 

=  sec~^a;. 


■■"s/a?—  1 


/-^ = -  sch-.. = -  log  fi + J^:^). 

^  x^ll~x^  \x       \a?       J 

r      do^  ^  _  ^^^^_,^ ^  _       A     JT^Y 

dx 

^2x  —  ^ 


C      dx  , 

I     .  =  vers~-^a;. 


36  INTEGEAL    CALCULUS.  [Art.  54. 

54.  We  took  up  in  I.  Chap.  V.  the  principal  devices  used  in 
preparing  a  function  for  integration  when  it  cannot  be  integrated 
directly. 

The  first  of  these  methods,  that  of  integration  by  substitution, 
is  simplified  by  the  use  of  the  difl'erential  notation,  because  the 
formula  for  change  of  variable  (I.  Art.  75  [1]), 

i  u=  I  uDyX    becoming     |  udx—  i  u  —  cly, 

reduces  to  an  identity''  and  is  no  longer  needed,  and  all  that  is 
requked  is  a  simple  substitution. 


rdx    , 

(a)   For  example,  let  us  find  I  —  VI  +  logaj. 

Let  1+ logic  =  2;, 

then  —  =  a2  ; 

X 

aud        J  -^  Vl  +  logaj  =J  zldz  =  %zl  =  f  (1  +  logo;)!. 


(&)  Required  j  - 


dx 


Let  e*  =  y, 

then  ^dx  —  dy, 

dx  e'dx  dy 


gx  _|_  g-x         g2x  ^  1  y2_^-^ 

and  f-T^^x  =  Tt^^  =  tan-i^/  =  tan.-V 

J  e^  +  e-"     J  l+y^  "^ 

(c)  Required  |  SQCx.dx. 


Let 

then 


seca;  = 

1 

coscc 

COSflJ 
cos  ^03 

z 

=  sin  a; 

> 

dz- 

=  cos  a;. 

dx. 

cos' 

'■x  =  \- 

-z\ 

Chap.  III.]  GENEEAL  METHODS   OF   INTEGRATING.  37 

i—  =  I :; i  =  ilog  -^— ,       by  Art.  53, 

cos ''a;       J  1  —  z^  1  —  z 

secic.aa;  =  flog 


1  —  sina; 
Examples. 


Prove  that  (1)  Ccscx.dx  =  ^log  ~ — ^^^  =  logtan  - 
J  1  +  cos  a)  2 


x'dx              1         T         a;  Vl  —  a;^ 
:=  —  ^cos~^a; 


Suggestion :  Let  x  =  cos  2;. 

55.    The   formula  for   integration  by  parts  (I.  Art.  79  [1]) 
becomes 

I  udv  =  uv  —  I  vdu,  -        [1] 

when  we  use  the  differential  notation.    It  is  used  as  in  I.  Chap.  V. 

(a)  For  example,  let  us  find  I  a;"  log  a;,  da;. 

Let  w  =  loga;, 

and  dv  =  x''dx\ 

then  du  =  — , 

x 


and  v  = 


«;• 


n  +  l 


n  +  l 


a;"loga;.aa;  = loga;  —  I dx  = f  los-o: 1. 

^  n  +  \     ^        J  n+l  n  +  l\    "^         n-{-l) 

(6)  Required  |  a;sin~^a;.dx. 

Let  if=sin~-'a;, 

and  dv  =  xdx ; 

then  du=      ^^     . 


INTEGRAL  CALCULUS.  [Art.  55. 


and  '"  ~  ^' 


I  xs,ixr^x.ax  =  —  sin    a;  —  -I- 1  —  , 

X  sin"-^  x.dx=^  —  sin~^  'x-\-\  (cos"^  x  +  x  V 1  —  oj^) . 


(c)  Eequired   i  — 


x^dx 


Let  w  =  a;e'', 

dec 


and  cZv 


then  du  =  (aje''  +  e"")  (^a;  =  e^(  1  +  a?)  c?aj, 

1 


and 


1+aj 
{\-VxY  1+a;     J  1  +  a;  1 


Examples. 

(1)  I —  —  =  sm^ — i-^^. 
-^  Vl-3x-a^  Vl3 

(2)  I  a;tan"-^a;.dx  =  — i^— tan~^a;  — -^-a;. 

^  ^  J  {1-xy^  ~  1-x    2{i-xy' 

(4)  I —  =  —  V2  ga;  —  a^  +  a  vers~-^-. 
»^  V2  aa;  —  3/*^  (^ 

(5)  rV2  aa;  -  a:^.  da;  =  ^  V2  ax  -  ar^  +  ^'  sin'^  '^^^. 
^  2  2  a 

Suggestion :  Throw  2  ax  —  a;^  into  the  form  a^  —  (x  —  a)^. 

(6)  I  -^ dx  =  log(x  -j-  sinx) . 

J  X  -{-  sinx 


Chap.  III.]    GENERAL  METHODS   OF   mTEGEATING.  39 

(/)     I — ■ c(a;=£ctan  — 

*/  1  +  cosic  2 

Suggestion :  Introduce  —  in  place  of  x. 

J  xijogxy^  ~  ~  {n~l)  (logcc)"-^" 

(9)  nog(logx)  ^^  ^  j^g^.  [log(loga;)  - 1] . 

(10)  (  ^  =  2;tan2;  +  logcos2;,  wliere2  =  sin~^cc. 

J  ( 1  —  or)  a 

(11)  r^^J^ =   1    logtan  f^  +  'L). 

J  sinaj+cosa;      Y2  \2      8/ 


40  INTEGKAL  CALCULUS.  [Art.  56. 


CHAPTER    IV. 

EATIOFAL    FRACTIONS. 

56.  We  shall  now  attempt  to  consider  sj'stematically  the 
methods  of  integrating  various  functions  ;  and  to  this  end  we 
shall  begin  with  rational  algebraic  expressions.  Any  rational 
algebraic  polynomial  can  be  integrated  immediately  hy  the  aid  of 
the  formula 

X" 


s- 


x^dx  = 


n  +  1 

Take  next  a  rational  fraction^  that  is,  a  fraction  whose  nu- 
merator and  denominator  are  rational  algebraic  polj'nomials. 
A  rational  fraction  is  proper  if  its  numerator  is  of  lower  degree 
than  its  denominator ;  improper  if  the  degree  of  the  numerator 
is  equal  to  or  greater  than  the  degree  of  the  denominator.  Since 
an  improper  fraction  can  alwaj^s  be  reduced  to  a  pol^'nomial 
plus  a  proper  fraction,  b}'  actually'  dividing  the  numerator  b}'  the 
denominator,  we  need  only  consider  the  treatment  of  proper 
fractious. 

57.  Every  proper  rational  fraction  can  be  reduced  to  the  siim 
of  a  set  of  simpler' fractions  each  of  which  has. a  constant  for  a 
numerator  and  some  poiver  of  a  binomial  for  its  denominator; 

that  is,  a  set  of  fractions  any  one  of  which  is  of  the  form  ■ 


fx 
Let  oui'  given  fraction  be  -^^  • 
^  Fx 


(x— a)' 


If  a,  &,  c,  &c.,  are  the  roots  of  the  equation, 

i^a;  =  0,  (1) 

we  have,  from  the  Theor}'  of  Equations, 

Fx  =A{x-  a)  (x  -  6)  {x  -  c) (2) 


Chap.  IV.]  RATIONAL  FRACTIONS.  41 

The  equation  (1)  may  have  some  equal  roots,  and  then  some  of 
the  factors  in  (2)  will  be  repeated.  Suppose  a  occurs  p  times 
as  a  root  of  (1),  b  occurs  q  times,  c  occurs  r  times,  &c., 

then  Fx  =  A{x-ay{x-by{x-cy (3) 

Call  A{x  —  by{x  —  cy--  =  (fiX', 

then  Fx  =  (x  —  ay^tx, 

fa  fa 

J.  ~  fx ;—  d>X  -. —  Ax 

and  ^  = ^ =  ^f^+       -"" 


Fx      {x  —  ay  cfix  {x  —  ay  (f)X      {x  —  ay  ^x 
fa  .       fa  ^ 

-r~  fx ; —  (bX 

4>a  •'        <^a  ^ 


fa 


{x  —  ay      {x  —  ay<^x 


f^~T^^^ 

—, is  a  new  proper  fraction,  but  it  can  be  reduced 

{x  —  ay(l>x  ^     ^ 

to  a  simpler  form  by  dividing  numerator  and  denominator  bj'^ 
a;  —  a,  which  is  an  exact  divisor  of  the  numerator  because  a  is  a 
root  of  the  equation 

fx—J—<^x=  0. 

If  we  represent  b}'  fiX  the  quotient  arising  from  the  division 
of  fx  —  ^  (fiX  by  X  —  a,  we  shall  have 

fa 
fx  ^        (jyq  fiX 

Fx       {x  —  ay      {x  —  a)"-^  cjiX ' 

•fry*  ' 

where ^ = —  is  a  proper  fraction,  and  may  be  treated 

{x  —  a)^~^  (jiX 

precisel}^  as  we  have  treated  the  original  fraction. 

fa 

Hence      i^ = il +  __.^^ . 

{x  —  ay-'^<l>x      {x  —  ay-'^      {x  —  ay-^(l)X 

By  continuing  this  process  we  shall  get 

fa  fa_  fa  f^.^a 

fx  ^       (i>a  f^a  4>a  .     <i>a        f^x 

Fx      {x  —  ay{x—ay-^{x  —  ay-'^  .x  —  a<i>x' 


42  INTEGEAL   CALCULUS.  [Art.  58. 

In  the  same  wa}'  -^  can  be  broken  up  into  a  set  of  fractions 

cf)X 

having  (x  —  b)^,  (a;  — 5)«~S  &c.,  for  denominators,  plus  a  frac- 
tion which  can  be  broken  up  into  fractions  having   (x—cy, 

(^x  —  cY'^ ,  &c.,  for  denominators;    and  we  shall  have,  in 

the  end, 

Fx      {x  —  ay      {x-  ay-^  '^  x-a      {x-by 

^'     ■         ^"  ■    ••  +  -£;  [1] 


{x  —  by~^  x  —  b 

where  K  is  the  quotient  obtained  when  we  divide  out  the  last 
factor  of  the  denominator,  and  is  consequent!}*  a  constant.  More 
than  this,  K  must  be  zero,  for  as  (1)  is  ideuticall}'  true,  it  must 

fx 
be  true  when  x=co  ;   but  when  x=  cc    I —  becomes  zero,  be- 

Fx 

cause  its  denominator  is  of  higher  degree  than  its  numerator, 
and  each  of  the  fractions  in  the  second  member  also  becomes 
zero;  whence  K=0. 

58.  Since  we  now  know  the  form  into  which  any  given  rational 
fraction  can  be  thrown,  we  can  determine  the  numerators  by  the 
aid  of  known  properties  of  an  identical  equation. 

Let  it  be  required  to  break  up -^-^ —  into  simpler 

iX  —  ^j"  \X  "i~  i- ) 

fractions. 

By  Art.  57, 

3a; -1  A        ,     B      ,     O 


{x-iy{x-^l)       (a;-l)-      x-1      aj  +  1' 

and  we  wish  to  determine  A,  B,  and  C.     Clearing  of  fractions, 
we  have 

3x-l=^A(x+l)  +  B(x-l)  {x+l)  +  C(x-iy.       (1) 

As  this  equation  is  identically  true,  the  coefficients  of  like 
powers  of  x  in  the  two  members  must  be  equal ;  and  we  have 

5  +  C=0, 
A-2C=3, 
A-B  +  C=-l; 


Chap.  IY.]  EATIONAL  FEACTIONS.  43 

whence  we  find  -4=1, 

C=-l; 
and  __3^^1 =_^^+J L..  (2) 

The  labor  of  determining  the  required  constants  can  often  be 
lessened  by  simple  algebraic  devices. 

For  example  ;  since  the  identical  equation  we  start  with  is 
true  for  all  values  of  x,  we  have  a  right  to  substitute  for  x  values 
that  will  make  terms  of  the  equation  disappear.  Take  equa- 
tion [1]  : 

3x-l  =  A{x+l)  +  B{x+l){x-l)  +  C{x-iy.      [1] 

Letx'  =  l,  2  =  2  A, 

A=l, 

then  2x-2  =  B  (x  +  1)  (x  -l)-\-C  {x-iy  ; 

divide  by  oj-l,  2  =  B  (^x  +  l)  +  C  (x-l). 

Leta;=l,  2  =  25, 

5=1, 

then  —x  +  l=C{x-l), 

C=-l. 

Examples. 

(1)  Show  that  when  we  equate  the  coeflficients  of  the  same 
powers  of  x  on  the  two  sides  of  our  identical  equation,  we  shall 
always  have  equations  enough  to  determine  all  our  required 
numerators. 

(2)  Break  up  — -^^ into  simpler  fractions. 

^  ^  ^    {x-3y{x-j-l)  ^ 

59.  The  partial  fractions  corresponding  to  any  given  factor 
of  the  denominator  can  be  determined  directly. 


44  INTEGRAL  CALCULUS.  [Akt.  59. 

Let  us  suppose  that  the  factor  in  question  is  of  the  first  degree 
and  occurs  but  once  ;  represent  it  b}^  x  —  a. 

=^=-^  +  ^  (1) 

Fx      X  —  a      (f>x''  ^  ^ 

by  Art.  57,  where 

Fx 

<jiX  = 


X  —  a 
so  tliat  Fx  =  (^x  —  a)  ^x. 

Clear  (1)  of  fractions. 

fx  =  A^x-\-{x  —  a)fiX.  (2) 

As  (1)  is  an  identical  equation,  (2)  will  be  true  for  any  value 
of  X.     Let  a;  =  a, 

fa  —  Affta, 

A  =  ^,  (3) 

a  result  agreeing  with  Art.  57. 

Hence,  to  find  the  numerator  of  the  fraction  corresponding  to 
a  factor  (x  —  a)  of  the  first  degree,  loe  have  merely  to  strike  out 
from  the  denominator  of  our  original  fraction  the  factor  in  ques- 
tion, and  then  substitute  a  for  x  in  the  result. 

If  the  factor  of  the  denominator  is  of  the  nth.  degree,  there  are 
n  partial  fractions  corresponding  to  it.  Let  {x  —  aY  be  the 
factor  in  question. 

fx  ^      A^               A^                  A         ,  ,     A     ■  ^^  .4x 

Fx     (x-ay     (^x-ay-^      (x-ay-^'^ '^x-a     cl>x^  ^ 

where  Fx  =  (x—aY  <l>x. 

fx 
Multiply  (4)  by  {x  —  a)",  and  represent  (a;  —  a)"  ^  by  f^x. 


^x=A^  +  A^ix  —  a)  +As{x  —  ay  + +  A^{x  —  a)""^ 

^x 


4.!A^(a;_a)". 


Chap.  IV.] 


EATIONAL   FRACTIONS. 


45 


Differentiate  successively  both  members  of  this  identity,  and  put 
x=a  after  differentiation,  and  we  get 

Ai  =  $a, 

2!  " 

Ai  =  —^>"a, 
3! 


A= 


(n-l) 


■  $("-^>a. 


Although  these  results  form  a  complete  solution  of  the  prob- 
lem, and  one  exceedingly  neat  in  theoryf  the  labor  of  getthig 
the  successive  derivatives  of  ^x  is  so  great  that  it  is  usually 
easier  in  practice  to  use  the  methods  of  Art.  58  when  we  have  to 
deal  with  factors  of  higher  degree  than  the  first.  So  far  as  the 
fractions  corresponding  to  factors  of  the  first  degi'ee  are  con- 
cerned, the  method  of  this  article  can  be  profitably  combined 
with  that  of  Art.  58. 


60.    As   an   example  where  the   method  of  the  last  article 
applies  well,  consider 


Sx-1 


=  ^+_A_+     ^ 


x(x-2){x  +  l)       X       x-2      ic  +  l' 
3a;-l 


A  = 

C  = 

Scc-l 


_{x-2){x-^l) 


^1 

1  =  0     2 


'_3£-l_"|     ^5 
x{x-\-l)_^^2     6 

'  3a;-l  " 

_x(x  —  2) 

=  i   1,5       1 


4 
3' 


a;(a;  — 2)(a;+l)      2  x      6   x-2      3   a^  +  l 


[1] 


46 


INTEGRAL   CALCULUS. 


[Art.  61. 


Again,  consider 

1                             1                     _        A                  B 

l  +  x^      (a;  +  V-l)  (a;-V-l)      iC+V-1      x-^T- 

-1 

A  — 

1 

1       v^ 

_X—'\J  —  \_ 

,  =  _vn        2V-1        2    ' 

B  — 

1 

1            V-i 

Lcf  + V-i_ 

^=^-1    2V-1           2 

1 

V-i 

1          V-i        1 

r 

l  +  aj2 


:+V-l. 


:-V-l 


[2] 


61.    Let  us  now  consider  a  more  difficult  example,  where  it  is 
worth  while  to  combine  our  methods. 

To  break  up ~ 

^   {x-iy{x^+l) 

a;3  +  1  =  (x  +  1)  (a^  -  a;  +  1) 
=  (a:+l)(x-i-|V^)(x-i  +  iV^3), 
a^  +  1  _  ^''  +  1  ^1 


{x-iy{x^+l)      {x-iy{x  +  l){ar-x+\)      {x-iy 
I        Ao        .        A^        .     Aj  B      . C 

^  (^x-iy    {x-iy'^x-i'^x+i'^^_^_x^z:s 


-i+iV,-3 


(1) 


B  = 

C  = 
D  = 


af+1 


{x-iy{x''-x  +  i) 

g;^  +  1 
_{x+l){x^  —  x+l)_ 

x^+1 


_  1 

:=-x"24' 

=  1, 


i{x-iy(x+l){x-l-^i^-3)j 
x^+1 


:=i+i,sAr3 


_{x  -  ly  (x  +  l){x-i-  i  V-3)>|-| v-3 


Chap.  IV.]  BATIONAL  FEACTIONS.  47 

_1  _1 

3  3  1     (2a;-l) 


Substitute  these  values  and  clear  (1)  effractions. 
24.{x^-{-l)  =  24.{x-{-l){x'-x+l)+24.A.Xx-l){x+l){x'-x-\-l) 
+  24.As(x-ly(x■j-l)(af-x+l)  +  24.A^(x-ly(x  +  l) 
(a^-.T+l)  +  (x-iyix'-x+l)  -8(2x--l)  (a;- 1)^^;  +1)  ; 

15x'^-olaf+A5x*+6x^-51x^+i5x-9=24:A2{x-l){x4-l) 
(a;2  _  ic  +  1)  +  24.43  (x  -  1)^  {x  +  1)  (a^  -  a;  +  1)  +  24^^ 
(x-iy(x  +  l)  (x^-^x-1). 

The  second  member  of  this  equation  is  divisible  by 
(^a;-l)(x+l)(x'-x-{-l), 
therefore  the  first  member  must  be  divisible  b}'  the  same  quantit3\ 
Dividing,  we  have 

15a;2- 3607  + 9  =  24^2  + 24^3(0; -l)+24^4(a;-l)2. 
Leta;=l,  — 12  =  24  Jla, 

2  2' 

and  we  get 

15^•2-36.^'  +  21  =  24^3(cc-l)  +  24^4(x-l)^ 

Divide  b}^  cc  —  1 ; 

15a;  -21  =  24^3  +  24^4(a;  - 1) . 
Leta.'=l,  —6  =  24^3, 

A  --^ 

^3  -  -  p 

15a;-15  =  24^4(a;-l). 
Divide  by  a;  —  1 ;  15  =  24  J.4, 


48  INTEGRAL   CALCULUS.  [Art.  62. 

Hence 

:^  +  l  _       1  1         1  1  1         .51 

(x-iy{x'  +  i)    {x-iy    2\x-iy    a' {x-iy    s'x-i 

+  1.J: i. '      _-l 1       _.      (2) 

24  x  +  1      3  rc-i-iV-S      3  a;-i  +  iV-3 


62.    Having  shown  that  any  rational  fraction  can  be  reduced 
to  a  sum  of  fractions  which  alwa^^s  come  under  one  of  the  two 

A  A 

forms  and  ,  it  remains  to  show  that  these  forms 

(x~aY  x—a 

can  be  integrated. 


To  find  f    ^^^ 

J  {x  —  a) 


let  z  =  x  —  a, 

then  dz  =  dx., 

and 


r    Adx      ^j^Cdz^  \         A^  1 A  PJ-, 


To  find  f  ^'^'^  , 
J  X  —  a 

let 

z  =  x—  a. 

then 

dz  =  dx, 

and  r-^^=^r— =  ^log2  =  ^log(a;-a).  [2] 

Turning  back  to  Art.  58  (2) ,  we  find 

/(3a;  — l)da;     _  r     dx  r  dx    _  C  dx    _  _     1 

{x-\y{x  +  l)~J  {x-iy     J  x-l     J  x+\~      x-1 

+  log(a;-l)-log(a;  +  l)  =  -- — -  +  log^^. 

Turning  to  Art.  60  (1),  we  have 

/{^x—\)dx     ^  jL  Cdx      5  r  dx     _  4  r  dx 
x{x-2){x-l)~^J   X       ^J  x-2       V  a;  +  l 

=  iloga;  +  f  log(a;  -  2)  -  Alog(x  +  1). 


Chap.  IV.]  RATIONAL    FRACTIONS.  49 

63.  If  imaginary  values  come  in  when  we  break  up  our  given 
fraction,  they  will  disappear  if  we  combine  oUr  results  properly 
after  integrating. 

We  know  (Art.  28,  Ex.  2)  that  if  the  denominator  of  our 
given  fraction  contains  an  imaginary  factor,  (x  —  a  —  b  V—  1)", 
it  will  also  contain  the  conjugate  of  that  factor,  namel}', 
(»  — a  + 6  V— 1)".  Moreover,  since  b^- Art.  59  the  numerator 
of  the  partial  fraction  corresponding  to  (a;  —  a  —  6  V  —  1)"  will  be 
the  same  rational  algebraic  function  of  «  +  6  V  —  1  that  the  nu- 
merator of  the  partial  fraction  corresponding  to  {x  —  a-{-b  V—  1 )" 
is  of  a  —  6  V  —  1,  these  two  numerators  must  be  conjugate  imagi- 
naries  by  Art.  28,  Ex.  3.      Hence,  for  every  fraction  of  the 

form  ^^t II — —  we  shall  have   a  second  of  the  form 

(a;  — a  — 5  V  — 1)" 


(cc  — a+&V  — 1)" 
J     A  +  bV^     ^,^^ 1 (A  +  B-J^l) 


f-, 


(.x--a-6.V-l)"  0^-1)    (ic-a-S  V— 1)"-^ 

by  Art.  62  [1], 

A-B^~l       ^i^^_       1  {A-B-J~\)      _ 


(x  — a  +  &V  — 1)"  O'' —  1)    (a;  — a  +  6V  — l) 

Let  (a;-c6  +  6V^)"-'  =  X+rV'^, 

X  and  Y  being  real  functions  of  x  ; 

then  (a;-a-&V^)"-'  =  X-rV^. 

A  +  B^^l        ,     ,r      A-Byf^l 


r    A  +  B^-^    ax+C-A 

^  (x  —  a—h^—iy  -J  (x  — 


dx 


(a;-a-6V-l)"  -^  (x  -  a  +  &  V-1)' 

^  1  {A  +  B  V^)  1  (A-B  V^) 

(n-l)'   X-r^^l        (n-1)      x+fV^ 

=  1  (2AX+2BY) 

~       (n-l)'  (.-c-  -  2  ax  +  a'  +  &')""'' 
a  result  which  is  free  from  imaginaries. 


[1] 


50  INTEGRAL  CALCULUS.  [Art.  63. 

If                                            n=l, 
we  have  the  pair  of  fractions, ■ 1:^:;  and 


x—a  —  hyJ  —  1         x—a+h'\/  —  l 

r  A  +  B-sJ-l    c^a;  =  (A  +  ^  V^)log(a;-a-6V3T). 

•^  cc  — a  — 6  V— 1 

by  Art.  62  [2], 

r_A--B^{^__ ^j^ ^  (^ _ 5  VHI)  iog(ic -  a  +  6 V^^)  ; 
^ x—a+b V— 1 

log(a;  -  a  -  6  V^)  =  ^log  [(«  -  ay  +  6^]  _  V^.  tan"^ .    ^ 


a;— a 


ic  — a 


log(aj-a  +  &V^)  =  ilog[(cc-a)2  +  &2]  +  V-l.tan-^—^ 

a;  — 

Hence        f-^+^V^  ,,^  +  r^-i?V^  ^, 

*^  x  —  a  —  b\/—l         ^  x  —  a  -\-by—l 

=  ^log[(iB-a)2  +  &2]  +  25tan-i-^,  [2] 

»  — a 

which  is  real. 

The  form  of  [2]  can  be  modified  by  adding  a  constant. 

"L  4.  tan-1  -^  -  -  4-  etn-^  ^"^^  -  !!:  _  etn-^  ^^-^'  -  tan-i  ^''  ~^ 
2  x-a      2  &         2  6  6 

Hence  Alog[(x-ay +  b''] +2Bt2in-^^^^:^  [3] 

differs  from   [2]  by  the  constant  jBtt,  and  therefore  is  a  true 

value  of     r_^+WHL<fa+  r_4zi^V^fe 

^  x-^a  —  b  -J—l  ^  X  —  a  +  byj—l 

Turning  back  to  Art.  61  (2)  we  find 

r      x^  +  i        ^^  ^  r    dx    _  1  r__dx__  _  1  r__dx__ 

,  ^  C  dx         ^    r  dx   _  j^  r dx j^  C dx 

^J  x-l      ^Va;  +  1      Va,_i_^V-3      V^^-i+iV^ 

=  -i I +  1 ^^ +  i._J_  +  |log(a;-l) 

3    {x-iy      4.    {x-iy      4.  x-\       ^     °^  ^ 

+  2\log(aj  +  l)-ilog(a;2-a;+l). 


Chap.  IV.]  KATIONAL  FEACTIONS.  51 

Examples. 

,.     r  oif  —  Sx  +  S      ,  ,  ^^x—2 

^  J  (x-l)  (x-2)  ^cc-1 

2)     I       ~    clx  =  X -\- ilos    ~   • 
■  ^  J  X-  +  1  ^x-i-2 

i        LtJi/  i  1 

I  — =  tan~^a;. 

J  x-^l 

C   dx         ,,       (x-iy  1    ,     _i2ic  +  l 

Jx^-l  ^a?  +  x+l       VS  V3 

C    dx           1    ,     _,  a;  ,     1    1      a+x 
I  — = tan^  -  -\ log  — !—  • 

J  a*  —  a;*       2a^  a      4:0?        a—x 

C 'h =  1  log^!±ii±l +J-  tan-2£+l, 

J  {x'  +  l){x''  +  x+l)  ^     x'  +  l        V3  V3 

r^M^_  ^  xiog^Zll  +  :^  tan-^^. 

Ja;4  +  a;2_2      "^     ^a;  +  l        3  V2 

— dx  =  hy-Og— ^• 

x^  +  x^  +  l  ^     ^x'+x  +  l 

C '^ = ^ ilog(a)-l) 

+  itan-^o;  -       ^^         +  ilog(a^  + 1) . 
4(03'' +  1) 

(10)  r.^!^^  j_w^-'^V2  +  i 

Jx^  +  l      4V2        a;2  +  ccV2  +  l 

+  _J_  rtan-XaJ  V2  + 1 )  4- tan-^iK  V2  - 1 )  ] . 
2V2 


52  INTEGRAL  CALCULUS.  [Art.  64. 


CHAPTER    V. 

KEDUCTION    FORMULAS. 

64.  The  method  given  in  the  last  chapter  for  the  integration 
of  rational  fractions  is  open  to  the  practical  objection  that  it  is 
often  exceeding!}^  laborious.  In  man}'  cases  much  of  the  labor 
can  be  saved  b}'  making  the  required  integration  depend  upon 
the  integration  of  a  simpler  form.  This  is  usuall}'  done  by  the 
aid  of  what  is  called  a  reduction  formula. 

Let  the  function  to  be  integrated  be  of  the  form  x^-'^{a-i-bx^y, 
where  m,  n,  and  p  may  be  positive  or  negative.  If  they  are  in- 
tegers, the  function  in  question  is  either  an  algebraic  polynomial 
or  a  rational  fraction;  if  they  are  fractions,  the  expression  is 
irrational.  The  formulas  we  shall  obtain  will  appl}-  to  either 
case. 

Denote  a  +  &.i'"  b}^  z  ;  then  we  want   |  x^~'^zP( 
Let  z^  =  u 


'dx. 


and  x'"~^dx  =  dv,  and  ijitegrate  by  parts, 

du  =  pz^~'^  dz  =  bnjjx"''^  z^~'^  dx, 
x"" 


m 


fx^-^z^dx  =  ^^  -  ^  Cx'^+'^-'^zP-^clx.  [1] 

This  formula  makes  our  integral  depend  upon  the  integi'al  of 
an  expression  like  the  given  one,  except  that  the  exponent  of  x 
has  been  increased  while  that  of'z  has  been  decreased. 

We  get  from  [1],  by  transposition, 

r^»  +  n-l  .^p-l  ^y.  ^  E!!!"  _  J!L    fa;™-!  gP  dx. 

J  bnp       bnpJ 


hence 


Chap.  V.]  EEDUCTION  FOEMULAS.  53 

Change  m  +  9i  into  m  and  p  —  1  into  p,  whence  m  is  changed 
into  m  —n  and  2^  into  j9  + 1 ,  and  we  get 

c^'r^-^.^dx = jL^r:^ _ ^^;-^\^  fa.— .--^c?..,   [2] 

J  6n(p+l)       6ft(p  +  l)J 

a  formula  that  lowers  the  exponent  of  x  while  it  raises  that  of  z.- 
Since  z=  a-\-  bx'^, 

Ta-^-i  2^  do;  =   fa;™-^  z^-'^  (a  +  bx'')dx  =  a  Cx"^- ^ z^-^  dx 

+  &  Cx'^-*-''-^zP-'^dx; 
therefore,  hy  [1], 

c^_  hnp  Crcr,.+n-i.^p-i^i^  ^  a  Cx"'-^zP-'^dx  -\-  b  Cx'^  +  '^-^zP-^dx, 
m        m  J  J  J 

Cx-H^-'  dx  =  ^'  _  H^  +  ^^P)    Cr,rr.^r.-l^,-laX. 

J  am  am       J 

Change  p  into  p  + 1  • 

Cx-H^dx  =  ^^^^^  -  Km  +  ^^j^  +  n)  r^™.n-i,.,;.,.      [-3] 
t/  c«7i  am  J 

Change  m  into  m  —  ?i,  and  transpose. 

C^^-H^ax^    .^-.^^^         a(m-n)    r^»-„-x,,,^^.       ^43 

We  have  seen  that 

faf^-^g^d-c  =  a  faj-^-^g^-^cZic  +  b  Cx'^  +  '^-H^-'^dx, 

and,  from  [1], 

b  Cx^  +  "-1  z^-^  dx  =  ^^  -  —  fee"-'  2^  cZa; ; 
J  np       npJ 


54  INTEGRAL   CALCTJLFS.  [Art.  65. 

hence 

/x'^-'^z^dx  =  a  Cx"'-^zP-'^dx  +  ^^  —  —  Cx'^-'^z^dx, 
J  np        npJ 

Cx'^-^znix^^^ — h    ^^P      Cx'^-'^zP-'dx.  [5] 

J  m  +  9ij3      m  +  npJ 

Change  p  into  p  +  1,  and  transpose. 

Cx^-'zPdx= ^^_^^  ^  m  +  np_±n  f^^-i^p  +  i^i^^      rg-i 

Formula  [3]  enables  us  to  raise,  and  formula  [4]  to  lower,  the 
exponent  of  x  by  n  without  affecting  the  exponent  of  z ;  while 
formula  [5]  enables  us  to  lower,  and  formula  [6]  to  raise,  the 
exponent  of  z  by  unit}-  without  affecting  the  exponent  of  x. 

Formulas  [1]  and  [3]  cannot  be  used  when  m  =  0  ; 

formulas    [2]  and  [6]  cannot  be  used  when  j9  =  —  1  ; 

formulas    [4]  and  [5]  cannot  be  used  when  m  =  —  Tip  ; 
for  in  all  these  cases  infinite  values  will  be  brought  into  the  sec- 
ond member  of  the  formula. 

65.    If  71=1,  z  =  a-\-bx, 

and  our  last  four  reduction  formulas  become 

cc^z^+i      b(m+p-\-l) 


j  x'^-^zHlx  = !^ —    ^  ^    ^  I  x'^zHx.  [31 

J  am  am  J  "-  -^ 

x'^-^zPdx  =  J- r  -  -) '-  I  x'^-HHx.  [4] 

o{m+p)      o{m-]-x))J 

Cx^-^zHx  =  '^^^    -\-~^^P—  Cx'^-^zP-^dx.  [5] 

J  m  -\-  p      m-\-  pJ 

{x'--^zHlx  =  -    ^"^"^^    _^m+j3  +  l    Cx^-iz^+^dx.        [6] 

If  7/1  and  p  are  integers,  and  m>0  and  j:>>0,  a  repeated  use 
of  [5]  will  reduce  p  to  zero,  and  we  shall  have  to  find  merely 

the  I  x'"-~^dx. 


Chap.  V.]  KEDUCTIOlsr  FOEMULAS.  55 

If  m<0  and  p>0,  [3]  will  enable  us  to  raise  m  to  0,  and 
then  [5]  will  enable  us  to  lower  j3  to  0,  and  we  shall  need 


onlyj^ 


If  m>0  and  p<0,  [6]  will  raise  p  to  —1,  and  [4]  will  then 

/dx 
— 
z 

If  m<0  andp<0,  [6]  will  raise  ^  to  —1,  and  [3]  will  raise 

iTdx 
m  to  0,  and  we  shall  need  1  — 
J  xz 


f 


m 


'dx      1 

—  =  log  a;, 

X 


fdx^  r 

J  z      J  a 

Cdx^  r 

J  xz     J  a. 


=  -\og(a  +  bx), 


+  bx      b 


dx 


1  ,      a  +  bx 
=  —  -  loo;  —^ 


x{a  +  bx)  a  x 

Hence,  when  w  =  1,  and  m  and  j)  are  integers,  our  reduction  for- 
mulas alwa3'^s  lead  to  the  desired  result. 


J  xr(a 


Examples. 
dx  b*i     a-\-bx  ,    W  b^      ,      b 


{a-]-bx)         a^  X         a}x      2a^xr      Sa-a^    4:  ax* 

(2)  Consider  the  case  where  n=2,  rewriting  the  reduction 
formulas  to  suit  the  case,  and  giving  an  exhaustive  investi- 
gation. 


^  ^  J  (a  +  bx-y^~ 


+ 


4:b{a-\-b3^y      8ab{a-\-bx^) 

tan'-'aj-v  -• 

8{ab)i  \a 


56  INTEGKAL   CAI.CULTJS.  [Art.  66. 


CHAPTER   VI. 

lEEATIONAL    FORMS. 

66.  We  have  seen  that  algebraic  polj'nomials  and  rational 
fractions  can  alwa3'S  be  integrated.  When  we  come  to  irrational 
expressions,  however,  ver^-  few  forms  are  integrable,  and  most 
of  these  have  to  be  rationalized  by  ingenious  substitutions. 

If  an  algebraic  function  is  irrational  because  of  the  presence 
of  an  expression  of  the  first  degree  under  the  radical  sign,  it  can 
be  easily  made  rational. 

Let  /(x,  "v/ct  +  bx)  be  the  function  in  question. 

Let  z  =  -\/a  +  bx  ; 

then  2"  =  «  -|-  bx. 

nz^~^dz  =  bdx, 

-,        nz'^-^dz 

dx  = ; 

6 

2"  —  a 
x  = 


b 

Hence        (f{x,  ^a  +  bx) dx  =  y  ( /[     ~^,  2  j«"'^ dz, 
which  is  rational  and  can  be  treated  b}-  the  methods  of  Chapter  IV. 

Examples. 

(1)  C^l^L±ldx  =  x  +  4:  ^x  + 4  log {^x-1). 
J  -^x  —1 

(2)  C^iax  +  bydx  =  ^^ V(«^'  +  bY+^^ 
J  a  (m  +  n) 

(3)  J{xy{x  +  a)  4-  V(^'  +  «)]f?^ 

2  u  +  1  n  +  1 


Chap.  VI.]  IRRATIONAL   FORMS.  57 

67.    A  case  not  unlike  the  last  is  j  f{x,  ^c  +  Va  +  bx)dx. 


Let  z  =  a/c  +  Va  +  6a; ; 


Hence 


z''  =  c  +  Va  +  bx, 
(s"— 0)™  =  a  -\-bx, 

(2"— c)™— « 

dx  = i-ii ^ 

|/(x%  "Vc  +  \/a  +  bx)dx 
mn   rS (z"  —  c)""  —  a     ~\.  „       x„_i  ,,_i 


-'^z''-'^dz. 


(1)  Find  r 

(2)  Find   f 


Examples. 

xdx 


Vc  +  V«  +  6a; 


dx 


A/l+Vl-a; 

68.  If  the  expression  under  the  radical  is  of  a  higher  degree 
than  the  first  the  function  cannot  in  general  be  rationalized. 
The  onl}'  important  exceptional  case  is  where  the  function  to  be 
integi'ated  is  irrational  bj'  reason  of  containing  the  square  root 
of  a  quantity  of  the  second  degree. 

Required     |  /(a;,  Va  -\-bx-\-  ca^)dx. 

First  Method.  Let  c  be  positive  ;  take  out  Vc  as  a  factor,  and 
the  radical  may  be  written  V^l  +  Bx  +  a^. 

Let  -y/A  +  Bx  +  a;-  =  a;  -|-  2, 

A  +  Bx  -\-  a^  =:  x^  +  2xz  +  z^, 

z^-A      . 

X  = , 

B-2z 


58  INTEGRAL   CALCULUS.  [Art.  68. 

a^^     2{z'-Bz  +  A)dz 
{B-2zy       ' 

J  A    .    T, — ; — 0  ,  z^  —  Bz  -\-A 

yA  -\-Bx  +  x'  =  x-\-z  = — -2— , 

B~2z 

and  the  substitution  of  these  vahies  will  render  the  given  func- 
tion rational. 

Second  Method.  Let  c  be  positive  ;  take  out  Vc  as  a  factor, 
and,  as  before,  the  radical  may  be  written  ^1 A  +  Bx  +  a?. 

Let  sfA+Bx^^^  = -sj A  +  xz  ; 

A  +Bx  -j-af  =  A-{-2  ^A .  xz  +  a?z^, 

2^A.z-B 
''-        \-z'       ' 

.       2{^A.z'-Bz  +  yJA)dz 
^A+Bx  +  x^=-^A  +  xz=^^-''-^'^'^^, 

1  —  Z" 

and  the  substitution  of  these  values  will  render  the  given  func- 
tion rational. 

If  c  is  negative  the  radical  can  be  reduced  to  the  form 
ylA+Bx  —  a^,  and  the  method  just  given  will  present  no 
difficulty. 

Third  Method.  Let  c  be  positive  ;  the  radical  will  reduce  to 
'sJA  -\-  Bx  +  XT.  Resolve  the  quantity  under  the  radical  into  the 
product  of  two  binomial  factors  {x  —  a)(x  —  (3),  a  and  fi  being 
the  roots  of  the  equation  A  +  Bx  -\-x^  =  0. 

Let  V(a;  —  a)  (x  — yS)  =  (x  —  a)z  ; 

(X  —  a)  (x  — /3)  =  (x  —  aYz^, 


X  =  ' 


z- 


a^^2z(f3-a)dz 
{1-zr    ' 

^/(x-a)(x-l3)  =  {x-a)z^  -^^^^^, 

1    ^ 


Chap.  VI.]  lERATlONAL  FOEMS.  59 

and  the  substitution  of  tliese  values  will  make  the  given  function 
rational. 

If  c  is  negative  the  radical  will  reduce  to  V^l  +  Bx  —  a^,  and 
may  be  written  ^f  {a  —  x)  {x  —  P)  where  a  and  ^  are  the  roots 
of  x^  —  Bx  —  xl  =  0,  and  the  method  just  explained  will  apply.. 

In  general,  that  one  of  the  three  methods  is  preferable  which 
will  avoid  introducing  imaginary  constants  ;  the  first,  if  c  >  0 ; 

a  a 

the  second,  if  c  <  0  and >  0  ;  the  third,  if  c  <  0  and  —  <  0. 

'  —  c  '  —  c 

a 
If  the  roots  a  and  (3  are  imaginary,  and  A  =  ^^  is  negative,  it 

will  be  impossible  to  avoid  imaginaries,  for  in  that  case 
A  +  Bx  —  x^  will  be  negative  for  all  real  values  of  x. 


69.    Let  us  compare  the  working  of  the  three  methods  just 

/clx 
' 

1st.    Let  '\/'2-{-ox  +  x^  =  x-{-z; 

r  dx  ^  r2(z--Sz  +  2)dz       3-22     ^  r  2dz 

J  -v/9-u::^^.4-^^'     J         (3-2zY        'z'-3z  +  2      J  S-2z 


f 


^2-{-dx  +  x^     ^         {S-2zr  2^-32  +  2 

=  -log(3-20), 

.  =  —  log(3  +  2  X  —  2  V2  +  3  ic  +  af') 


dx 


V2  +  3  a;  +  ar 

T  1 

=  log 


3  +  2  a.'  —  2  V2  +  3  a;  +  a;^ 


_,  3+2a;  +  2V2+3a;  +  a^' 

~  ^°9  +  12a;  +  4.^•^-8  -12a;-4a;2 


=  log[3+2a;  +  2V2  +  3a;  +  a;2].  (1) 

2d.    Let  V2  +  3  a;  +  a;-  =  V2  +  a;z  ; 

r  dx  ^  2  r(V2.^--3g  + V2)c?2  ^  1-z^ 

J  V2  +  3a;  +  ar^       J  (1-^')'  ^2  .z'-Sz-{-'^2 

=  2  C-^  =  log^.  (Art.  52) 

J  1  —  Z'  1  —  z 


60  ■  INTEGRAL   CALCULUS.  [Art.  69. 


dx  ^  j^^.  x-^2  +  V¥-\-3x  +  :x^ 

's/2  +  'dx  +  x^         °  x  +  ^2  —  ^2+3x-{-x^ 


^»  x'+2-^2.x  +  2-2-nx-a^ 

\      S  +  2x  +  2^2  +  'dx-^x^ 
=  ^"S 2Vi^3 


=  log(3  +  2x+2^2+-dx+x^)  -  log(2V2-3), 
or,  di'opping  the  constant  log (2  a/2  —  3) , 

r  ^^^       —  =  log(3  +  2a;  +  2V2  +  3a;  +  a^)-  (2) 

•^  V2  +  3  a;  +  a;- 

3d.    Let  V2  4-3iK  +  ar^=  V(a^+1)  {x +  2)  =  (x  +  l)z ', 


J 's/2  +  Sx-\-x'        J  {1-zy     -z  Jl-z^  1- 


z 


J  -.n. 


dx 


1+     f^+2 


^•^  +  1  _  i^^Va;+l  + Va;  +  2 


:  =  log ,'  =  log 


V2  +  3a;  +  a.'^         °l_     !•£+_£         '^  Va;+ 1  —  Va; +2 
\a;  +  l 

,      CK  +  1  +  2  V2  +  3  a;  +  ic^  _^  a;  _|_  2 

=  loa;  — ■ ■ ■ ■ ■ — 

^  a;  +  l-a;-2 

=  log  (3  +  2a^  +  2  V2  +  3a;+a^)  +  log  (-  1), 

or,  dropping  the  imaginarj"  constant  log  (  —  1 ) , 

f  ^^^       — =log(3  +  2a;  +  2V2  +  3a;  +  a.-^;).  (3) 

-^  V2  +  3x  +  «2  ' 

Examples. 

/dx  n  V4  +  2  ic  —  V2  —  X 
^^=r  =  4  log — =1=^ • 
(2  +  3a;)V4-x2              V4  +  2a;+V2-a; 

(2^     r  /^       =  log  (I-  +  X  +  Va^  +  a.0 . 
J  V  a;-  +  a; 

(3)    r  ,        '^"^         -  =  —  log f—  +  x^/c  +Va  +  to  +  ca.-^\ 
^     ■  J  Va  +  to  +  cx-      Vc        \2Vc  / 


Chap.  VI.]  IRRATIONAL   FORMS.  61 

70.  If  the  function  is  irrational  tlirough  tlie  presence,  under 
the  radical  sign,  of  a  fraction  whose  numerator  and  denominator 
are  of  the  first  degree,  it  can  alwa^'s  be  rationalized. 


Required  i/f  .r,  -T  "  "*"    )  clx. 
J     \      \lx-\-m 


Let  z  =  ^'^^^±^ 

yilx+  m 


+  m 
^„  _ax-\-b 
Ix  -\-  m 

X  — , 

Iz''  —  a 

I ,_  njcim  —  bl)z''~^dz 

and  the  substitution  of  these  values  will  make  the  given  function 
rational. 

Example. 

r     dx        sll-x ^ _  3  3 1/1 -a^Y 


71.  If  the  function  to  be  integrated  is  of  the  formx"'~\a-j-bx^y, 
m,  ri,  and  j9  being  an}'  numbers  positive  or  negative,  and  one  at 
least  of  them  being  fractional,  the  reduction  formulas  of  Art.  64 
will  often  lead  to  the  desired  integral. 


Examples. 
=  fsin  ^x. (3  +  2ar), 


{l-x^)h 


dx  11      1  -  Vl  -  x^      Vl 


ar'Vi_a;2  x  2ar' 

J  {2ax  —  x-')i  \2       2/  \2a 

C4^   f     xhlx      ^      {2a^  +  Sx^) 


62  INTEGRAL   CALCULUS.  [Art.  72, 

72.  We  have  said  that  when  an  irrational  function  contains  a 
quantit}"  of  a  higher  degi-ee  than  the  second,  under  the  radical 
sign,  it  cannot  ordinarily  be  integrated.  It  would  be  more  cor- 
rect to  saj-  that  its  integral  cannot  ordinarily  be  finitely  expressed 
in  terms  of  the  functions  with  which  we  are  familiar. 

The  integrals  of  a  large  class  of  such  irrational  expressions 
have  been  speciall}-  studied  under  the  name  of  Elliptic  Functions. 
The}-  have  peculiar  properties,  and  can  be  expressed  in  terms  of 
ordinar}^  functions  only  by  the  aid  of  infinite  series. 


Chap.  VII.]         TRANSCENDENTAL  FUNCTIONS.  6'^ 


CHAPTER  VII. 

TRANSCENDENTAL    FUNCTIONS. 

73 .   In  dealing  with  the  integration  of  transcendental  functions 
the  method  of  integration  by  parts  is  general!}"  the  most  effective. 

For  example.     Required   |  a;  (log cc)- dec. 

Let  M  =  (loga7)^, 

dv  =  x.dx ; 

'2,\ogx.dx 
du  = » 

X 

a? 

p  (log  x)  2  =  ^""(^^^^y  _  Cx  log  x.dx  =  -  [  (log  x)  2  -  log  a;  +  i] . 

Again.     Required   i  e'sm.x.dx. 
u  =  sin  X, 
dv  =  e"  dx ; 
dti  =  cos  x.dx, 

I  e^'sin  x.dx  =  e^^sin  a?  —  I  e'cosx.dx, 

I  e"  cos  x.dx  =  e'  cos  x-\-  i  e"  sin  x.dx ; 

,                       Ct  '        1        e''(sina;  — coso;) 
whence  I  e*  sm  x.dx  =  -^ ' , 

,  Ct  1        e*(sincc +  cosx) 

and  I  e""  cos  a;.ax'=  — ^^ ■ '— 

J  2 


64  INTEGRAL,   CALCULUS.  [Akt.  74. 

Examples. 


(1)      x^'^ilogxydx  =  — — (log.'c)^-    ^    »/ 


6  loo;  a;  6 


(?ft  +  1)-      (m  +  1)3 

^  ^  J  (l-a;)2       1-a;  ^"^  ^ 

(3)    fe"^ V(l  —  e-'^).dx  =  J-e«==  V(l  —  6)^'"+  isin-^e<^. 

74.  The  method  of  integration  b}^  parts  gives  us  important 
reduction  formulas  for  transcendental  functions.  Let  us  con- 
sider I  sin^x.dx. 

u  =  sin"~^a;, 
dv  =  sinx.dx; 

du  =  {n  —  l)sin'^~^xcosx.dx, 
V  =  ~  cos  x ; 

I  siu"a.'.cZa;  =  —  sin"~^  ic  cos  cc  +  (n  —  1)  |  sin"~-a;  eos'X.dx 
=  — sin"~^cccos"iK  +  (?i  —  1)  I  (sin"~^a;— sin"a;)da; ; 

/s'lrf-x.dx  = sin"~^x'  cosa^  +  ^^~     )  sin"+^a;.da;.  [1] 

n  n   J 

Transposing,  and  changing  n  into  n  +  2,  we  get 

/I                                         7^  _f_9    /• 
sin" x.dx= sin" + ^ cc  cos  a;  H ^^^  i  sin" '^'^x.dx.       r2 1 
1i-t-l                                   71+1  J  "-   -^ 

In  like  manner  we  get 

/ cos" x\c7ic=  -since cos""' a^H ^^—  |  cos"~^a;.da;,  [3] 

n  n    J 

/cos'^ x.dx  =  —  — — ■sina.'cos"+'a; +  ^^— t^  j  cos"  +  ^a;.cZa;.  r4'l 
71  +  1  n  +  lJ  ^  -^ 

If  n  is  a  positive  integer,  foi^mulas  [1]  and  [3]  will  enable  us 
to  reduce  the  exponent  of  the  sine  or  cosine  to  one  or  to  zero, 


Chap.  VII.]        TEANSCEKDENTAL  FUNCTIONS.  65 

and  then  we  can  integrate  by  inspection.  If  n  is  a  negative 
integer,  formulas  [2]  and  [4]  will  enable  us  to  raise  the  ex- 
ponent to  zero  or  to  minus  one.    In  the  latter  case  we  shall  need 

,  or   I  — — ,  which  have  been  found  in  Art.  54  (c) . 

coscc        J  since 


Examples. 


3 

■X. 


(1)  Jsin^a;.dx  =  -  ^^^Y^^^Ain^g;  +1)  +  : 

/o\     C      &     J        since cos^a^/      «      ,  5\   ,    5  /  .  ,     ^ 

(2)  I  cos''a;.aa;  = (  cos"a;  +  -    H (smcccosce+a;). 

^  ^  J  6         V  4y      16  ^  ^ 

(3)  r-^  =  -H^?^+ilogtanf. 

(4)  Obtain  the  formulas 

fsh"  x.dx  =  -  Sh»-ia;  Chcc  -  ^^—1  C^lx^-'^x.dx.  [11 

J  n  71    J  -^ 

rSh"a;.da;  =  ^— Sh"+ia;Chcc-^i±^  C^'\x''+^x.dx.    [21 
J  n+l  n  +  lj  ■-  -■ 

fch" x.dx  =  -^\xx Ch"-i X  +  ^^^  fch^-'ce.da;.  [3] 

fch" x.dx  ■-= ^—  Sh X  Ch" + ^  a;  +  ^^  fch" + ^ x.dx.  [41 

c/  9i+l  w+lJ  ^  -■ 

^  ^  J  Sh'^o."  'Sh-x'      *    ^^ChcK+l 

75.    The  (sin"^cc)"daj  can  be  integrated  by  the  aid  of  a  reduc- 
tion formula. 

Let  z  —  sin~^cc ; 

then  cc  =  sin2:, 

dx^=  cosz.dz, 
and  I  (sin~^a;)''da;=  j  2"cos2.d2. 


6Q  INTEGRAL  CALCULUS.         [Art.  76. 

Let  ?« =  2", 

dv  =  cosz.dz ; 
du  =  nz^'^dz, 
v  =  smz; 

i  z"cosz.dz  =  z'"smz  —  ni  z"~^sinz.dz. 

i  z^'^sinz.dz  can  be  reduced  in  the  same  waj^,  and  is  equal 
to  —z^~^cosz-\-{n  —  l)  i  z'^~^ cosz.dz; 
hence 

i  z"" cosz.dz  =  z"smz  +  'nz'"~'^ cosz  —  n(n  —  l)  ( z"~^cosz.dz,    [1] 

or  I  (sin~-^a;)"da;  =  cc(sin~^a;)"+  izVl  —  a3^(sin~^cu)"~^ 

—  n(n  —  l)  r(sin-^a;)"-^da;.  [2] 

If  w  is  a  positive  integer,  this  will  enable  us  to  make  our  re- 
quired integral  depend  upon  |  dx  or  |  sin~^x.dx,  the  latter  of 
which  forms  has  been  found  in  (I.  Art.  81). 

Examples. 

(1)  Obtain  a  formula  for  |  (vers"^a;)"da;. 

(2)  r(sin-^a;)*da;  =  a;[(sin-iaj)*-  4.3.  (sin-^a^)2+4  .3.2.1] 

+  4  Vl  -  a^sin-^o;  [(sin-^aj)2-  3 .  2]. 

76.  Integratioh  by  substitution  is  sometimes  a  valuable  method 
in  dealing  with  transcendental  forms,  and  in  the  case  of  the  trigo- 
nometric functions  often  enables  us  to  reduce  the  given  form  to 

an  algebraic  one.     Let  it  be  required  to  find  |  (/;sina;)  cosx.dx. 

Let  2;  =  sinx, 

dz  =  cosx.dx; 

I  (fsmx)  cosx.dx  =  ifz.dz. 


Chap.  VII.]         TRANSCENDENTAL   FUNCTIONS.  67 

In  the  same  way  we  see  that 
I  {fcosx)smx.dx  =—  ifz.dz  if  z  =  cobx^ 

and 
j  [/(since,  cos  cT)]  cos  a.  da;  =      i  [/(g,  \J\—z^)'\dz  if  2  =  sin  a;, 

I  [/(sinx,  cosa;)]  sina;.c?x  =  —  |  \_f{z,  '\Jl—z^)']dz  if  2:  =  cosa:. 

77.     I  sin'^a;  cos^x.da;  can  be  readily  found  by  the  method  of 

Art.  76  if  m  and  n  are  positive  integers,  and  if  either  of  them 
is  odd.     Let  n  be  odd,  then 

71-1 

cos"  a;  =  cos""\r  cos  a;  =  (1  —  sin^a;)  2  cos  a;, 
I  sin™ a;  cos"a;.cLT  =  |  sin™ a;  (1  —  sin'a;)^"  cosaj.da;. 

Let  2  =  sin  a;, 

dz  =  co&x.dx, 

sin™  a.' cos"  a;. da;  =  I  2™(1  —  2;^)  2  ^z^ 

which  can  be  expanded  into  an  algebraic  polynomial  and  inte- 
grated directty. 

If  m  and  n  are  positive  integers,  and  are  both  even, 

I  sin™  a;  cos"a;.cZa;  =  |  sin™a;(l  —  sin^a;)2'da;. 

n 

sin™a;(l  —  sin^a;)2"  can  be  expanded  and  thus  integrated  by 
Art.  74  [1]. 

If  m  or  n  is  negative,  and  odd,  we  can  write 

cos"  X  =  cos"~^  X  cos  X,     or     sin™  x  =  sin™"^  x  sin  a;, 

and  reduce  the  function  to  be  integrated  to  a  rational  fraction 
b}'  the  substitution  of 

2;  =  cosa;,     or    2:  =  sina;. 

I  sin™  a;  cos"  a;. da;  can  also  be  treated  by  the  aid  of  reduction 
formulas  easily  obtained. 


68  INTEGEAL  CALCULUS.  [Art.  77, 


I  cos^  X  Vsin  x.dx-= 


Examples. 
,    _  cos^^cc      cos®  a 


10  8 

2  sini  cc  _  2  sin J  x 

~~i  7 


rsin^^2cosi^_2c^g.^, 
»^     Vcosa;  5 

/,      -47        sin  a;  cos  cc /sin*  .T      sin^cc      1\   ,    x 
cos-o.  sm^x.do;  = ^—  (^-^^ 12-  -  8 j  +  16 

—  =  seca?  +log  tan  — 

sin  X  cos-  X  2 

/dx  cos  a:;  ,  3 ,  ,  x 
=  sec  X — \-  -  log  tan- 
sin^  a;  cos^  a?                   2  sin"  a;      2  2 

/'  dx                 1,1,,. 
= — I -— — f- log  sin  a;, 
tan^a;         4  tan*  a;      2  tan'' a; 


Chap.  VIII.]  DEFINITE  INTEGRALS.  69 


CHAPTER    VIII. 

DEFINITE    INTEGRALS. 

78.  A  definite  integral  has  been  defined  as  the  limit  of  a  sum 
of  infinitesimals,  and  we  have  proved  that  if  the  function  to  be 
integrated  is  continuous  between  the  values  between  which  the 
sum  is  to  be  taken,  this  limit  can  be  found  by  taking  the  differ- 
ence between  two  values  of  an  indefinite  integral. 

In  some  cases  it  is  possible  to  find  the  value  of  a  definite 
integral  from  elementary  considerations  without  using  the  indef- 
inite integral,  and  it  is  worth  while  to  take  one  or  two  examples 
where  this  can  be  done. 

JIT 
cos^x.dx. 

By  our  definition  this  must  equal 

limit [cos^O.daj  +  co&^ dx.dx -\-  cos^2 dx.dx  -{■  cos^Sdx.dx  -\- 

-f-  cos^(Tr—ddx) .  dx  +  cos^(7r—  2 dx) .  dx  +  cos^(7r— da;) .  dx'] 

=  limit  [^dx  +  cos^  dx.dx  +  cos^2dx.dx  +  cos^Sdx.dx  -j- 

—  cos^3  dx.dx  —  cos^  2  dx.dx  —  cos^  dx.dx^ , 

since  cos  (tt  —  ^)  =  —  cos  <^. 

We  see  that  in  this  sum  the  terms  destroy  each  other  in  pairs, 
with  the  exception  of  the  first  term  dx  if  the  number  of  terms 
is  odd,  and  with  the  exception  of  the  first  term  and  a  term 

cos^- dx.dx  in  the  middle  of  the  set,  if  n  is  even.     Each  of  these 

2 
terms  has  zero  for  its  limit  as  dx  approaches  zero  ;  hence 

cos^x.dx  =  0. 


/' 


70  mTEGKAL   CALCULUS.  [Art.  79. 

sirr'x.dx. 

XTT 
sin^x.dx 

=  limfsiu^O.cZrK  +  sin^dx.dx  +  siTo?2dx.dx  +  sin^Sdx.dx  + 

+  sin^  (tt  —  3  dx) .  dx  +  sin^  (tt  —  2  da;) .  dx  +  sin^  (tt  —  dx) .  da;] 

=  lim[0  +  2s[n^dx.dx  +  2  sin^  2  dx.dx-\- 2  sin^  3  dx.dx -\- ] 

=  2  1  ^sin-a;.da;  =  21ini[sin^0.da;  + siii^cZa;.cZa;+sin^2cZa;.da;+ 

+  sin^ (  -  —  2 dx \.dx  +  sin^ (-  —  dx\. dx"] 

=  21ini[sin^0.cto  +  sin^cZa;.da;  +  sin^2da;.da; +  sin^3da;.da;  + 

+  cos^  3  dx.dx  +  cos^  2  dx.dx  +  cos^  da;,  da;] 
=  2  lim  [da;  +  da;  +  da;  + ] . 

Since  sin^  <^  +  cos^  <A  =  1 , 

but  dx -\- dx -\- dx -\- =-. 

4 

Hence  |  sin- a;.da;  =  —  = -. 

Jo  4        2 

Examples. 

I  cos^aj. da;    and 

0 

sin- a;. da;  by  the  usual  methods. 

J27r 
sin^a;.da;. 

80.  It  is  generally  necessaiy,  however,  to  obtain  a  required 
definite  integral  by  substituting  in  the  value  of  the  indefinite 
integral  according  to  Art.  78,  and  this  can  always  be  done  when 
the  function  to  be  integrated  is  finite  and  continuous  between  the 
values  between  which  the  definite  integral  is  to  be  taken  ;  that  is, 
between  what  are  called  the  limits  of  integration. 


Chap.  VIII.]  DEFINITE  INTEGRALS.  71 

Examples. 
1)     p5H^.  Ans.    V2-1. 


x^  2 


^)  Jo    V^ 

5)  I  e'^^dx  (a  positive). 

6)  Ce-'^smmx.dx.  ^ws. 

J  ••CO  yy 

0  a  + 


^ws.   — 
a 

m 
a 


81.  "When  we  have  occasion  to  use  a  reduction  formula  in 
finding  a  definite  integral,  it  is  often  worth  while  to  substitute  the 
limits  of  integration  in  the  general  formula  before  attempting  to 
find  the  indefinite  integral. 

—  . 

We  can  reduce  the  exponent  of  x  by  Art.  64  [4] 

J  b{m  +  7ip)      h{m  +  np)J  ^  -" 

For  our  example  this  becomes 

Cx^-\cv'-x'')-hdx 

-m+1  -m+lJ         ^  ^ 

When  a;  =  0,  ^^ —  =  0,    and  also  when  x  =  a. 

—  m  +  1 


72 
Hence 


INTEGRAL   CALCULIJS. 


[Art.  81. 


~^dx. 


n™-i  (a^  _  x')  -i  dx  =      ^-!^ — ^  Cx^-Hci'-  x")  "I 
Jo  m  —  1   Jo 

Cx'  (a"  -af)-idx      =      —  Cx'  {a'  -  x')  'i 

=      a'^  i  x-(a^—x^)~i 

6  4     Jo     ^  ^ 

= + ^  §  1  a6  r   ^^^ 

q'a'2     Jo   V^^2^ 

/dx  .  _iX 

—  =  sm    -1 

—  =    sin"^-        —    sin"-^-        =  sin~-'(l)  — sin~-'(0). 

sin~^(l)  =  -    or    -  +  27r    or    -H-47r.     In  areneral,  -  +  2 wir. 
V  /      2  2  2  2 


idx 
'idx 


sin~^(0)  =  0    or    27r  or    4-77. 


In  general,  2w7r. 


If  we  take  2?i7r  as  the  value  sin  ^  has  when  x=0,  and  then 

increase  a?  to  1,  as  all  the  increments  of  the  sin~^  ——=r^   are 

Va^— ar 
positive,  our  whole  increment  must  be  positive,  and  we  must  take 

2  WTT  +  -  as  the  value  of  sin~^  (1 ) .     Hence 


sin"^(l)  — sin~^0  =  -     and 
^  ^  2 


0  Va' 


x^dx 


13      5    TTtt^ 


-^      2  4   6    2 


Examples. 


J^°    ar^dT         2  4 
0  ^/^,^^v2-3■5' 


Ct    ~^  tC   »LtJu  — — 


(3)     )["'«' Vo^I^. 


ttO,'' 


,      _  1     TT^ 

^^~  4.'  4  ■ 


Chap.  VIII.]  DEFINITE   INTEGRALS.  73 

(4 )     (  af(a^  —  ocr)^^  dx 

/-N    oi        ^1    4.   r^  •  «      7       1.3.5 (w  — l)7r 

(o)   Show  that  I     sm"x.cto= — ,.,    ,  ■ ^-wheni 

Jo  2.4.6 n      2 


1    3_^ 

q'iq 


171  IS  even 


2.4.6 (n-1)     ,  .       ,, 

= ^^ ^  when  n  is  odd. 

3.5.7 n 


I     cos"cc.cZx=  I     sin"  a;,  dec, 
,     C^   x-"-dx 


^  a-''"c?a;    _  1.3.5 (29t  — 1)  tt 

2.4.6 2n       2 

(Suggestion :  Let  x  =  sin  6.) 


82.    Required  i  e'^'a^cZa;. 

I  e'" x"  dx  =  —  e"' x"" -\- n  |  e~''a;""V?a;,  by  integration  by  parts. 

When  a:  =  0,  —  =  0;  but  when  a  =  oo,  —  =  — ,  and  is  inde- 
ed e""      00 

terminate.     Determining  it  b}'  the  method  of  I.  Art.  141,  we 
find  tliat  its  true  vahie  is  0  ;  hence 

e~''x''dx  =  n  |  e'^^ic^'^do;  =  ?i(w  —  1)  |  e"''a;"~^da;.  (1) 

If  ?i  is  an  integer,  this  gives  us 

I  e~^x''dx  =  nl  j  e~''dx. 
0  Jo 

/e'^dx  =  —  e""'  = ,  which  is  equal  to  0  when  a;  =  oo,  and 

to  —  1  when  x  =  0, 

e~''dx  =  1, 

If  n  is  not  an  integer  the  vahie  of  the  definite  integral  is  much 

e'^x'^'^dx 

is  generally  represented  by  r  (?i) ,  and  it  has  been  carefully 
studied  under  the  name  of  the  Gamma  Function. 


74  INTEGRAL   CALCULUS.  [Art.  83. 

In  every  case  we  have  by  (1), 

C  {logxydx  can  be  reduced  to  (  —  1)"  (  e'^z'^dz,  and  there- 
fore =  ±r(?i  +  l). 

83.    We  have  seen  that  if  fx  becomes  infinite  for  a  value  of  a; 
between  a  and  &, 

I  fx.dx  is  not  equal  to      i  fx.clx        —     ifx.dx 

J'^c-E  /^i 

fx.dx  and    1  fx.dx  can 

be  found  without  difficult}'. 

limit  r  f^'^  r^         ~\ 

^A    I   fx.dx -^  I  fx.dx    is  called  the  p^'incipal  value  of 

I  fx.dcfi,  and  is  often  finite  and  determinate. 

— ■ — 

a    C  —  X 

=  00  when  x=c. 

c  —  x 

""^  dx        1     c—  a 


n-'  dx        , 
I      =  log 

i/a        C  —  X 

r^    dx   _ 

Jc  +  E  C  —  X 


,      h  —  c 
loo- , 


and  the  principal  value  of  i    — '- —  =  log 

J  a  C  —  X  h  —  c , 

Example. 


— -• 
x„  x^ 


Ans.  ^(—^ K\- 

xi      X 


84.    We  have  seen,  Art.  51,  that  a  definite  integral  is  a  func- 
tion of  the  limits  of  integration,  and  not  of  the  variable  explicitly 


Chap.  VIII.]  - 


DEFINITE  INTEGEALS. 


75 


appearing  in  the  expression  integrated.     Let  us  consider  the 
possibility  of  differentiating  a  definite  integral. 

Requu'ed  Da  I  f{x,a)dx  where  a  is  independent  of  x,  and 

oya 

a  and  b  do  not  depend  upon  a.     Of  course  our  derivative  is  a 

partial  derivative. 


Let       u  ■ 


-  I  f{x,a)dx, 


DaU  = 


limit 
Aa  =  0 


X'5  /^h 

f(x,a-{-  Aa)dx—  I  f(x,a)dx 


Aa 


limit   r  f 
^a=0LJ„ 


'/(a;,a  +  Aa)-/(a;,a)^^; 


Da  i  f{x^a)dx  =  I  Daf(x,a)dx, 
and  we  find  that  we  can  differentiate  under  the  sign  of  integration. 


The  truth  of  the  converse  of  the  last  proposition  can  be 
easily  established. 


dx 
dx  if  a,   6, 


I  \  f{x^a)dx\da.—  I         \  f{x^a)da 

\  f{x^a)dx    da  =   I  I     f(x,a)da 

ao,  and  a  are  entu'el}'  independent. 

A  skilful  use  of  the  principles  of  the   present   Article    will 
often  enable  us  to  obtain  new  definite  integrals. 

Jc^  1 

I  e  '^dx  =  — 
0  a 

Differentiate  both  members  with  respect  to  a  ; 

{  — xe~'" dx) ^=. ;    or       I  {xe~'"dx-=—- 

Differentiate  again ; 


^re-^'dx. 


2! 


76  INTEGRAL  CALCULUS.  [Art.  85. 


Differentiating  n  times ; 


Let  us  consider   ' 


X 


e~'"dx  =  — 


a^+a      2   as 
Differentiate  n  times  with  respect  to  a  ; 

TT    1.3.5 (271-1)      1 


r"^      dx       ^ 

Jo  (x^  +  ay+^~ 


2.4.6 2n        ce+h 


85.    By  differentiating  under  the  sign  of  integration  a  compli- 
cated form  may  sometimes  be  reduced  to  a  simpler  one. 

e'""  sin  mx.dx 


Requu'ed  | 

Let      u=  I     dx, 

Jo  X 


D, 


e'^'^cos  mx.dx  =  — -,       by  Art.  80,  Ex."  7 

0  a^  +  m^ 


Hence      u=  \  —-^ — -dm  =  a  i    „^^^^  „  —  tan~^—  +  C. 
J  a^  +  tiv  J  a^  +  m^  a 

Since,  when  m  =  0,  dx  is  constantly  zero, 

X 

dx  =  0  when  m  =  0,  and  therefore  C  =  0,  and 

0  X 

we  have  ^^  _^  . 

i     e  "^  sin  mx  ,        ,       ,  ??i 

I   ' dx  =  tan"^  — 

Jo  X  a        ■ 


Example. 


J<*\                 1 
x''dx= 
0                 n-\-l 

J}\^ogxfdx={-lf-^^^^^^^^ 


Chap.  VIII.]  DEFINITE  INTEGRALS.  77 

Jfi  1 

0  a 


Integrate  both  members 

r(  rx-^cia)cu=  p^, 

c/O  lOffCC  On 


Examples. 


I     e"""  cos  mx.  clx  =  — -.  obtam 

0  a^  +  nv 

0  X  \a^  —  m?) 

e-"^  sin  mx.dx  =  — obtaui 

0  a-  +  m- 

X H — sm mx.dx  =  tan^  -^  —  tan  ^-^• 

0  x  m  m 

87.    If  in  Cf(x,a)dx  a  and  &  are  functions  of  a,  our  formula 
for  —  I  /(x,a)fZx  becomes  more  complicated. 

Let  ffi^^o.)  ^^^  =  -P'Ca^^a) » 

then  u=  ff{x,a)dx  =  F(b,a)-F{a,a), 

'h  =  £F(b,a)-^F{a,a)', 

da       da  da 

but  as  &  and  a  are  functions  of  a, 

—  F(b,a)=D,F{b,a)'^  +  DaF{b,a), 
da  da 

±F(a,a)  =  D,F{a,a)'^-{-DaF(a,a),       by  I.  Art.  200. 
da  da 


78  INTEGRAL  CALCULUS.  [Art.  88. 

A^(&,a)=/(6,a), 
D,F{a,a)=fia,a); 

^  =  Da[F{b,a)  -  Fa,a^  +/(&,«)  f  - /(a,a)  ^* 
cla  Cla  tta 

=  CDaf{x,a)dx)  +  f(b,a)^—  -f{a,a)  ^-^.  [1] 

t/o  Cla  ua 

Example. 

Coufirm  this  formula  bj'  finding  —  I  sin  (x  +  y)  dx. 

dyJ'o 

,  88.  In  our  treatment  of  definite  integrals  we  have  supposed 
that  our  limits  of  integration  were  real,  and  that  the  increment 
dx  of  the  independent  variable  was  always  real. 

Definite  integrals  taken  between  imaginary'  limits,  and  formed 
by  giving  the  variable  imaginary  increments,  have  been  made  the 
subject  of  careful  stud}',  and  they  are  found  ver}'  useful  in  con- 
nection with  the  subject  of  Elliptic  Integrcds  (Art.  72),  or,  as 
the}'  are  sometimes  called.  Doubly  Periodic  Fimctions. 


Chap.  IX.]  LENGTHS  OF  CUEVES.  79 


tanr : 

~  dx 

and  (I. 

Arts. 

52  and  181^ 

1  that 
ds'-. 

=  da?  +  dy^ 

From  these 

we  get 

sinr: 

dy 

~  ds' 

COST 

_dx 
ds 

CHAPTER  IX. 

LENGTHS    OF    CURVES. 

89.  If  we  use  rectangular  coordinates,  we  have  seen  (I.  Art.  27) 

[1] 

[2] 
[3] 

[4] 

by  the  aid  of  a  little  elementar}-  Trigonometr}' . 

These  formulas  are  of  great  importance  in  dealing  with  all 
properties  of  cm-ves  that  concern  in  nny  waj^  the  lengths  of  arcs. 

We  have  already  considered  the  use  of  [2]  in  the  first  volume 
of  the  Calculus,  and  we  have  worked  several  examples  by  its 
aid  in  rectification  of  cm-ves.  Before  going  on  to  more  of  the 
same  sort  we  shall  find  it  worth  while  to  obtain  the  equations  of 
two  very  interesting  transcendental  curves,  the  catenary  and  the 
tractrix. 

The  Catenary. 

90.  The  common  catenary  is  the  curve  in  which  a  uniform 
heav}'  flexible  string  hangs  when  its  ends  are  supported. 

As  the  string  is  flexible,  the  onl}^  force  exerted  by  one  portion 
of  the  string  on  an  adjacent  portion  is  a  pull  along  the  string, 
which  we  shall  call  the  tension  of  the  string,  and  shall  represent 
by  T.  T  of  course  has  different  values  at  different  points  of  the 
string,  and  is  some  function  of  the  coordinates  of  the  point  in 
question. 


80 


INTEGEAL   CALCULUS. 


[Art.  90. 


The  tension  at  an}'  point  has  to  support  the  weight  of  the  por- 
tion of  the  string  below  the  point,  and  a  certain  amount  of  side 
pull,  due  to  the  fact  that  the  string  would  hang  vertically  were 
it  not  that  its  ends  are  forcibly  held  apart. 

Let  the  origin  be  taken  at  the  lowest  point  of  the  curve,  and 

suppose  the  string  fastened 
at  that  point. 

Let   s  be  the   arc   OP, 
P  being  an}-  point  of  the 
string.  As  the  string  is  uni- 
form, the  weight  of  OP  is 
proportional  to  its  length ; 
-^  we  shall  call  this  weight  ms. 
This  weight   acts  verti- 
callj'  downward,  and  must  be  balanced  by  the  vertical  effect  of  T, 
which,  \)-y  I.  Art.  112,  is  Tsinr. 

Hence  'TsinT  =  ms.  (1) 

As  there  is  no  external  horizontal  force  acting,  the  horizontal 
effect  of  the  tension  at  one  end  of  an}'  portion  of  the  string  must 
be  the  same  as  the  horizontal  effect  at  the  other  end.  In  other 
words,  TcosT  =  c  (2) 

where  c  is  a  constant.     Dividing  (1)  b}^  (2)  we  get 

s  =  —  tanr, 
m 

or  s  =  atanT,  (3) 

where  a  is  some  ^constant.    From  this  we  want  to  get  an  equa- 
tion in  terms  of  x  and  y. 


tanr  =  Vsec-T  —  1  =  a  hr^  —  1  ; 
\  ax- 


hence 


or 


and 


s^  =  cr 


rW 


-  1 


a?ds^ 
ads 


^dx' 
=  {a'-\-s')dx^, 

=  dx. 


Integrate  both  members. 


Chap.  IX.]  LENGTHS   OF  CURVES.  81 

a  log(s  +  Va^  +  S-)  =  x-\-Ci 
when  cc=  0,  5  =  0, 
hence  (7  =  log  a, 

^ 

and  log  (s  +  Va2+  s-)  =  -  +  log  a, 

/ - 

ya^+s^  =aea  —  s, 

X  X 

s  =  -(e«  — e  «)  =  atanr       by  (3). 
Hence  a^=  -  (e«- e"!), 

and  2/  =  ^(e"  +  e"«)  +  C. 

If  we  change  our  axes,  taking  the  origin  at  a  point  a  units 
below  the  lowest  point  of  the  cuiwe,  y  =  a  when  x  =  0,  and 
therefore  (7=0,  and  we  get,  as  the  equation  of  the  catenary, 

y  =  |(e"  +  e~«).  (4) 

Example. 

Find  the  curve  in  which  the  cables  of  a  suspension-bridge 
must  hang.  Ans.   A  parabola, 

Tlie  Tractrix. 

91.  If  two  particles  are  attached  to  a  string,  and  rest  on  a 
rough  horizontal  plane,  and  one,  starting  with  the  string  stretched, 
moves  in  a  straight  line  at  right  angles  with  the  initial  position 
of  the  string,  dragging  the  other  particle  after  it,  the  path  of  the 
second  particle  is  called  the  tractrix. 

Take  as  the  axis  of  X  the  path  of  the  first  particle,  and  as 
the  axis  of  Y  the  initial  position  of  the  string,  and  let  a  be 


82 


INTEGRA!.  CALCULUS. 


[Art.  9L 


the  length  of  the  string.     From  the  nature  of  the  curve  the 
string  is  always  a  tangent,  and  we  shall  have  for  any  point  P 


y 

-  =  —  sinr, 

for  T  lying  in  the  fourth  quadrant  has  a  negative  sine, 
r 


[1] 


t 


=  sm^T 


dy^ 


df 


hence 


and 


a"  ds^      daf  +  dy^ 

y^{da?  +  d'lf)  =  a^dy^, 

2/2  da?  =  (a^  —  y^)  dy^, 
{a^-yy^dy 


dx=±- 


y 


is  the  differential  equation  oi^etractrix. 

On  the  right-hand  half  of  the  curve  t  is  in  the  foui'th  quadrant, 

-^  or  tanr  is  negative,  and  we  shaU  write  the  equation 
dx 


dx  =  — 


{o?-yy^dy 

y 


[2] 


If  we  allow  the  radical  to  be  ambiguous  in  sign  we  shall  get 
also  the  curve  that  would  be  described  if  the  first  particle  went 
to  the  left  instead  of  to  the  right.  The  tractrix  curve,  generally 
considered,  includes  these  two  portions. 

Integrating  both  members  of  [2] ,  and  determining  the  arbi- 
trary constant,  we  get 

x  =  -  ^ce-f  +  glog^"^^^'"^'  [3] 

as  the  equation  of  the  tractrix. 


Chap.  IX.]  LENGTHS   OF   CUEVES.  06 

Examples. 

(1)  Show  by  Art.  91  (1)  that  in  the  tractrix  s  =  a  log-  if  s  is 
measured  from  the  starting-point.  ^ 

(2)  Find  the  evolute  of  the  tractrix.     (I.  Art.  93.) 

Rectification  of  Curves. 

92.  In  finding  the  length  of  an  arc  of  a  given  curve  we  can 
regard  it  as  the  limit  of  the  sum  of  the  difierentials  of  the  arc, 
and  express  it  hy  a  definite  integral. 


We  shall  have 


Of  course  in  using  this  formula  we  must  express  Vc?a;-  +  cly^ 
in  terms  of  x  only,  or  of  y  only,  or  of  some  single  variable  on 
which  X  and  y  depend,  before  we  can  integrate. 

For  example  ;  let  us  find  the  length  of  an  arc  of  the  circle 

XT  -{-  y-  =  a^. 
2x.dx-\-2y.dy  =  0, 


dy  = 


y 


/y*2      I        ni"  r^Z  rytZ 

da^  +  dy^  =     '^y  dx?  =  -  dxr  =  — ■  dx^, 

y^  2/  <^      ^ 

s  =  al    —  =  a  sm  ^— —  sm^— )• 

c/«o  -s/a^-af        \         a  a  J 

The  length  of  a  quadrant  =  a  (  sin  -  =  —  ; 

Jo       Va^  -  x'       2 

.*.  the  lenoth  of  a  circumference  =  27ra. 


84  INTEGRAL  CALCULUS.  [Art.  93. 

Length  of  Arc  of  Cycloid. 

93.    For  the  cj'cloid  we  have 

x  =  a$  —  a  sin  < 
y  =  a   —a  cos  ( 
dx  =  a{l  —  cos$)d9  =  ydd, 

^  =  vers"^^, 
a 


(I.  Art.  99.) 


d6  =  ± 


1         dy  dy 


\    a 


y^      -si  2  ay  — y'^ 


a" 


dx  =  —   ^  y       1 
V2  ay  —  f' 

ds^=dx^-^dy^=  ^''y^y\==  ^^^^\ 

2ay  —  y       2  a  — y 

ds^sJYa.       ^y       . 
V2a  —  y 

s  =  -J¥a  P  ,  ^y       =  2V2~a(V2^"^o  -  V2^r=70- 
•^*o  V  2  a  —  2/ 

Kthe  are  is  measured  from  the  cusp,  ^/o  =  0, 

s  =  4a  — 2  V2aV2a  —  2/i.  [1] 

K  the  arc  is  measured  to  the  highest  point,  yi  =  2  a, 

s  =  4a. 
The  whole  arcTi  =  8  a. 

Example. 

Taking  the  origin  at  the  vertex,  and  taking  the  direction  down- 
ward as  the  positive  direction  for  y,  the  equations  become 

^  =  ciO  + a  sine)  (I.  Art.  100.) 

y=   a  — a  cos  0  ) 

Show  that  s  =  2  V  2  ay  when  the  arc  is  measured  from  the 
aunmit  of  the  curve. 


Chap.  IX.] 


LENGTHS   OF   CURVES. 


85 


and 

s  =  a  ^2 


94.   "We  can  rectify  the  cycloid  without  eliminating  $. 
x  =  a6  —  a  sin  I 
y  =  a  —  a  cos  I 
dx  =  a(l  —  cos  ^)  dO, , 
dy  =  a  sin 6. dO J 
dx"  +  df=2  aHO'O-  -  cos^), 


2sin2^ 
2 


s=:aJ2C  (l-cosOy-dd, 

Jo, 

^'d6  =  4a  j^ sin ^f?|  =  4a  fcos^j  -  cos|\ 


If  ^0  =  0  and  ^1  =  2  TT,  we  get  s  =  8a  as  the  whole  curve. 


95.   Let  us  find  the  length  of  an  arch  of  the  epicycloid. 


x  =  (a-\-  h)  cosO  —  b  cos^  '6 

0 

y=  (a  -{-  b)  sin  ^  —  6  sin     ~l    0 


,(I.Art.l09[l].) 


dx  = 
dy  = 


■  (o  +  b)  sin  ^  +  (a  +  6)  sin^^^-t-^^ 
(a  +  b)  cos^  -  (a  +  b)  cos^^-i^^ 


dO, 
dd. 


as^  =  (a  +  bydO-  2-2  /"cos^^^^  cos^  +  sin^^i-^^  sin^^j 
=  2(a  +  &)'cZ^'('l  -  cos^^Y 


s  =  (a  +  6)V2  f  \l- co&^e]dd, 


_4&(a  +  5) 


a  /J  a  /J 

cos  —  ^0  —  cos  —  Pi 
26  26 


[1] 


2& 


To  get  a  complete  arch  we  must  let  6q=Q  and  ^i  =  —  tt. 

Hence,  for  a  whole  arch, 

86(a  +  6) 


86 


INTEGRAL   CALCULUS. 
Examples. 


[Art.  96. 


(1)  Find  the  length  of  an  arch  of  a  hypocycloid. 

(2)  Find  an  arc  of  the  curve  xi  -j-  yi  =  a?,  and  see  whether  it 
agrees  with  the  result  of  Ex.  (1) ;  see  I.  Art.  109,  Ex.  2. 


96.    Let  us  attempt  to  find  the  length  of  an  arc  of  the  ellipse. 

f 
IP- 


— I- —  =  1 


a 


2x.dx      ^y.dy 


1j    ^ 

cly  =  --^  dx, 
a-y 


a^f 


1  + 


Ipy? 


=r[- 


&2  ™2 
XT 


a?{c^  —  a^) 
''dx. 


dv?. 


9/9  9\ 

a-  (cc  —  xr) 
This  function  cannot  be  integrated  directly ; 


1  + 


62  a^ 


a\a^-x^_ 

can  be  expanded  b}'  the  Binomial  Theorem,  and  the  terms  can  be 
integrated  separate^,  and  we  shall  have  the  length  of  the  arc 
expressed  by  a  complicated  series. 

A  more  convenient  way  of  dealing  with  the  problem  is  to  use 

an  auxiliary  angle.  Instead  of  -3  +  p  =  1  we  can  use  the  pair 
of  equations 

X  =  a  sin  ^ 
y  =  b  cos  (f> 
dx  =  a  cos  (fi.d(ji, 
dy  —  —  b  sin  (f>.dcf), 
d^  =  (a^cos^ <^  +  b'sm^<j>)d(lP  =  [ci^-  (a^  -  b')  sin^ <f>']dcf>^ 

=  cr-fl  -  ^!!^sinVy?<^'  =  a\l-  e'sm'cf,)dcf>\ 
where  e  is  the  eccentricity  of  the  ellipse. 


}' 


(I.  Art.  150), 


Chap.  IX.]  LENGTHS  OF  CUKVES.  87 

s  —  ai  {l  —  e^sm^cf>)hd<f> 

XI       2  ^24  ^     2   4   6  r      J   '^ 

This  is  one  of  the  famous  Elliptic  Integrals. 

Example. 
Show  that  the  length  of  the  hyperbolic  arc  is 


.=.£[i+fs.'. 


'd<fi. 


Polar  Formidae. 

97.  If  we  use  polar  coordinates  we  have 

ds  =  Vd^^+r^,    (I.  Art.  207,  Ex.  2) , 
tane  =  '-|^,  (I.  Ai-t.207). 

0/7* 

From  these  we  get,  by  Trigonometry, 

rddi  rn-i 

sm.  =  -J,  [3] 

cos  €  =  — .  r41 

ds  -^ 

98.  Let  us  find  the  equation  of  the  curve  which  crosses  all  its 
radii  vectores  at  the  same  angle.     Here 

tane  =  a,     a  constant, 

rd(fi  _ 
——  —  a, 
dr 

adr  _  ,, 

r 

alogr  =  (^  4-  (7, 


INTEGRAL  CALCULUS.  [Art.  99. 


0  0    a 


r  =  be<'  (1) 

where  &  is  some  constant  depending  upon  the  position  of  the 
origin.  This  curve  is  known  as  the  Logarithmic  or  Equiangular 
Spiral. 

99.  To  rectify  the  Logarithmic  Spiral.   We  have,  from  98  (1), 

a        ^b 

ad>  =  a — J 
r 

rd<f>=  adr, 
d^  =  dr^-{-  r^dcf>^  =  (1  +  a^)dr^ ; 

s  =  f(l  +  a')idr  =  (1+  a')h(r,  -  n). 

Examples. 

(1)  Find  the  length  of  an  arc  of  the  parabola  from  its  polar 
equation 

1  +  cos  <^ 

(2)  Find  the  length  of  an  arc  of  the  Spiral  of  Archimedes 

r  =  a(ji. 

100.  To  rectify  the  Oardioide.     "We  have 

r=2a(l-cos<^),  (I.  Art.  109,  Ex.  1), 

dr  =  2asmcf).dcfi, 

d^z=  4a-sin^<^.d^^  4-4a^(l  —  cos^)^d<^^ 
=  8a^d^^(l  — cos<jf)), 

s  =  2^2.a  C(l-coscf>)hdc^=8a   cos^^-cos^' 
•J'Po  [_       2  2_ 

=  1 6  a  for  the  whole  perimeter. 


Chap.  IX.] 


LENGTHS   OF    CUKVES. 


89 


Involutes^' 

101.  If  we  can  express  the  length  of  the  arc  of  a  given  curve, 
measured  from  a  fixed  point,  in  terms  of  the  coordinates  of  its 
variable  extremity-,  we  can  find  the  equation  of  the  involute  of 
the  curve. 

We  have  found  the  equations  of  the  evolute  of  y=fx  in  the 
form 

x'  =  X  —  pCOSv 

y'  =  y  —  psmv 
We  have  proved  that     tanv  =  tanr', 

dp 

nT'  =  ^ 

ds 

dx' 


]' 


and  that 


1, 


COST   = 


ds'} 


(I.  Art.  91). 

(I.  Ai-t.  95), 
(I.  Art.  96)  ; 

(Art.  89). 


Since         tanv  =  tanT',    v  =  t'   or   v=180°  +  t'. 

As  normal  and  radius  of  curvature  have  opposite  directions, 
we  shall  consider  v  =  180°  +  r'. 


Then 
Hence 


smi/  =  —  smr 


X    =:X-{-  p- 


y'  =  y+pTj' 


and     cos  v  =  —  cos  t' 
dx' 

dy^ 
'ds'' 
Since  dp  =  ds', 

P  =  s'  +  l 

where  I  is  an  arbitrary  constant,  x  and  y  being  the  coordinates 
of  any  point  of  the  involute,  it  is  only  necessary  to  eliminate  x', 
y',  and  p  hy  combining  equations  (1) ,  (2),  and  (3)  with  the  equa- 
tion of  the  evolute. 

As  we  are  supposed  to  start  with  the  equation  of  the  evolute 
and  work  towards  the  equation  of  the  involute,  it  will  be  more 
natural  to  accent  the  letters  belonging  to  the  latter  curve  instead 


(1) 
(2) 

(3) 


90 


INTEGRAL   CALCTJLITS. 


[Art.  101. 


of  those  going  with  the  former ;    and  our  equations  may  be 

written  ^  _  ^.'  _i_  „'  ^' . 

els  ' 


X  =x'-\-  p'"-^ 


(4) 


y  =  y'-\-p 


<(iy 


cls 


(5) 

p'=s+L  (6) 

To  test  our  method,  let  us  find  the  involute  of  the  curve 


f  = 


21m 


{x  —  mY, 


(7) 


letting  p'=s-\-m.     We  must  first  find  s. 

Q 

2  ydy  = {x  —  mydx, 

9m 


_   4    (a;  —  my 
9  m        y 


dy  =  ^  '-'"      ""'  dx, 
2x-\-m 


ds'  = 


3  m 


dx^, 


V3 


/Q'  =  5  +  m  = 


1       /^^  1 

—   I  (2a;  +  m)5dx'  =  — (2a;  +  m)i— m, 


■  (2a:;  +  m)i, 


,  ,  2x  -\-m 

x  =  x  -\ ■ — 

3 


y  =  y'+ 


3^3m 
-m 

4     (2a;  +  ffl-)  (a?  —  m)^ 


27m 


2/ 


,      x  —  m 
X  = «-, 


4  ,           ,2          4a;'2 
2/  =  -  —  (a;  -  m)2  = , 

92/  2/ 

£c  =  3a;'+m, 
4a;'2 


2/  =  -- 


2/' 


Substituting  in  (7)  the  values  of  x  and  y  just  obtained,  we  have 
y'^  =  2mx' 
as  the  equations  of  the  required  involute. 


Chap.  IX. J  LENGTHS  OF  CXJIIVES.  91 

Example. 
Find  the  involute  of  ay^  =  x^. 

.102.   The  involute  of  the  cycloid  is  easily  found.     Take  equa- 
tions I.  Art.  100  ((7), 

x=   aO  -\-  a  sin  6 
y  =  —a  +  a  cos  6 

p'  =  S, 

a 

dx=-a{l-\-cos6)d9      =2acos^-clO, 

a  a 

dy  =  —  asin0.d$  =  — 2asin-cos-c?^, 

2       2 

ds^  =  2  a^  dO''  ( 1  +  cos  ^)  =  4  a2  dO'"  cos^  ^, 


Let 


=^"X 


cos-dd        =4asm-? 


iC  =  aj'  +  4asin-cos-  = ;«'  + 2a  sin^, 

2        2  ' 

?/  =  ?/'  + 4a  sin^-         =  ?/'  — 2a(l  — cos^), 

x'  =  aO  —  a  sin  ^ 
y'  =z   a  —  a  cos  ^ 

a  cycloid  with  its  cusp  at  the  summit  of  the  given  cycloid. 

Example. 
From  the  equations  of  a  circle 

X  =  a  cos  (f> 
y  =  a  sin  (^ 
obtain  the  equations  of  the  involute  of  the  circle. 

Ans.   x'=  a  (cos  ^  +  ^  sin  <^)  ) 
y'=  a(sin  (fi  —  (fi  cos cji)  3 


92  INTEGEAL  CALCULUS.         [Art.  103. 

Intrinsic  Equation  of  a  Curve. 

103.  An  equation  connecting  the  length  of  the  arc,  measured 
from  a  fixed  point  of  any  curve  to  a  variable  point,  with  the 
angle  between  the  tangent  at  the  fixed  point  and  the  tangent  at 
the  variable  point,  is  the  intrinsic  equation  of  the  curve.  If  the 
fixed  point  is  the  origin  and  the  fixed  tangent  the  axis  of  X,  the 
variables  in  the  intrinsic  equation  are  s  and  r. 

We  have  already-  such  an  equation  for  the  catenary 

s  =  atanr,  Art. 90  (3),     [1] 

the  origin  being  the  lowest  point  of  the  curve. 
The  intrinsic  equation  of  a  circle  is  obviously 

s  =  ar,  [2] 

whatever  origin  we  may  take. 

The  intrinsic  equation  of  the  tractrix  is  easily  obtained.     We 

have 

y  =  —  a  sinr,  Art.  91  (1) , 

and  s  =  alog-;  Art.  91,  Ex.  1, 

hence  s  =  a  log  ( —  esc  t) 

where  t  is  measured  from  the  axis  of  X,  and  s  is  measured  from 
the  point  where  the  curve  crosses  the  axis  of  Y.  As  the  curve  is 
tangent  to  the  axis  of  Y,  we  must  replace  t  by  t  —  90°,  and  we 

^^  s  =  logsecT        ^c<i(K$JU.r-      [3] 

as  the  intrinsic  equation  of  the  tractrix. 

Example. 

Show  that  the  intrinsic  equation  of  an  inverted  c^'cloid,  when 

the  vertex  is  origin,  is 

s  =  4asinT;  (1) 

when  the  cusp  is  origin,  is 

s  =  4a(l— cost).  (2) 


Chap.  IX.]  LENGTHS   OF   CURVES.  93 

104.    To  find  the  intrinsic  equation  of  the  epic3'cloid  we  can 
use  the  results  obtained  in  Art.  95. 

da;=(a  +  6)(^sin^^^-sin^V^=2(a+&)cos^^±^^sin^^.d^, 
\         b  J  2b  2b 

dy=={a-\-b)fcos6-eos^-!^o\w=2{a+b)sm^-!^^esm^e.de, 

by  the  formulas  of  Trigonometr}' ; 

sin  a  —  sin/3  =  2  cos ^ (a  +^8)  sin|(a  — /?), 

cos/8—  cosa  =  2  sin|(a  +  fi)  sin^(a  —  (3)  ; 

,  ch/      .      a  +  2b^ 

tanr  =  ^i  =  tan  — ■ 0, 

dx  26 

hence  t  =  — rr^O; 

2b 


-i^(^>(l-cosA^^  byArt.95[l]; 


therefore  s  =  iM^L±A)  (i  _  cos      ''      r]  [1] 

a         \  a-\-2b  J  "-  "^ 

is  the  intrinsic  equation  of  the  epic^'cloid,  with  the  citsjy  as  origin. 
If  we  take  the  origin  at  a  vertex  instead  of  at  a  cusp 

a 
7r(a  +  25)        ,. 

T f-T    , 

2a 

hence  s'  =  — ^  '  sin t'  ; 

a  a  +  2b 

4b(a-\-b)    .         a 

or  s  =  — ^^ — ■ — ^sin T 

a  a  +  2  6 

is  the  intrinsic  equation  of  an  epicycloid  referred  to  a  vertex. 


94  INTEGEAL   CALCULUS.  [Art.  105. 

Example. 
Obtain  the  intrinsic  equation  of  the  hypoej^cloid  in  the  forms 

45(a— 6)/-,                a 
s  —  — ^^ ^   1  —  cos 


«-26    '•  <^>- 


4  6 (a— 6)   .        a  .^. 

s  =  — 5^ ^sin T.  (2) 

a  a-2b  ^  ^ 


105.  The  intrinsic  equation  of  the  Logarithmic  Spiral  is  found 
without  difficulty. 

We  have  r=&e«,  (Art.  98), 

and  s=  Vl+a2(ri-ro).  (Art.  99). 

If  we  measure  the  arc  from  the  point  where  the  spiral  crosses 
the  initial  line,  ro  =  b,  aud  we  have 

^ 

s  =  6Vl  +a-(e"-l). 

In  polar  coordinates  t  =  ^  +  e,  and  in  this  case  e  =  tan~^  a  ;  if 
we  measure  our  angle  from  the  tangent  at  the  beginning  of  the 
arc  we  must  subtract  e  from  the  value  just  given,  and  we  have 

or,  more  briefly,        s  =  k{c^  —1),  k  and  c  being  constants. 

106.  If  we  wish  to  get  the  intrinsic  equation  of  a  curve  directly 
from  the  equation  in  rectangular  coordinates,  the  following  method 
wUl  serve : 

Let  the  axis  of  X  be  tangent  to  the  curve  at  the  point  we  take 
as  origin. 

tanT  =  ^;  (1) 

ax 

and  as  the  equation  of  the  curve  enables  us  to  express  y  in  terms 
of  x^  (1)  will  give  us  x  in  terms  of  t,  say  x  =  Ft', 


Chap.  IX.] 
then 

hence 


LENGTHS   OF   CURVES. 
dx  =  F't.cIt,  ♦ 


dx 


dx 


=  F't-,    but    —  =  cost; 
ds  ds  ds 

ds  =  sec  tF't. dr. 


95 
divide  hy  ds ; 

(2) 


Integrating  both  members  we  shall  have  the  requked  intrinsic 
equation. 

For  example,  let  us  take  a^=  2  my,  which  is  tangent  to  the 
axis  of  X  at  the  origin. 


2xdx  = 

2  mdy, 

dx 

tanT  = 

X 

m 

dx  = 

m  sec^- 

r.dr, 

dx  _ 
ds 

COSt  = 

m  sec^ 

T  — 

ds  =  m  sec^T.dr, 


^ 


(1) 


r   dr         m[  sinr     ,  ,       ,       /tt   ,  t\]   ,    ^ 
=  m    — -=-  — ^+logtan  -+-     +(7, 
J  cos"^T      2  |_cos^T  \4      2J_  ■ 

=  0;         .-.   (7=0; 
f  logtan^^  +  I^  . 


s  =  0  when  t  =  0  ; 

sinr 


COS^T 


(2) 


Examples. 

(1)  Devise  a  method  when  the  curve  is  tangent  to  the  axis 
of  F,  and  applj'  it  to  2/^  =  2  mx. 

Q 

(2)  Obtain  the  intrinsic  equation  of  ?/^  = {x  —  my. 

21  m 

(3)  Obtain  the  intrinsic  equation  of  the  involute  of  a  circle. 
(Art.  102,  Ex.) 


96 


INTEGRAL,   CALCULUS. 


[Art.  107. 


107.    The  evohde  or  the  involute  of  a  curve  is  easily  found 
from  its  intrinsic  equation. 


If  the  curvature  of  tlie  given  curve  decreases  as  we  pass  along 
the  curve,  p  increases,  and 

s'  z=  p  —  po.  (I.  Art.  96) . 

If  the  curvature  increases,  p  decreases,  and 
s'  =  po  —  p. 
Hence  always  s'  =  ±(p  —  po)  j  [1] 

ds 


P  = 


dr 


(I.  Arts.  86  and  90). 


We  see  from  the  figure  that  t'  =  t. 


Hence 


s'  =  ± 


\dTjr=T'       \drjT=o_ 


or,  as  we  shall  write  it  for  brevit}", 

,  ds 
s  =  ±  — 
d 


[2] 


108.    The  evolute  of  the  tractrix  s  =  a  log  sec  t  is 
d  loo;  sec  - 


dr 


—  atanr,     the  catenary. 


The  evolute  of  the  circle  s  =  ar  is 

d 


s  =  a- 


di 


=  0,     a  point. 


Chap.  IX.]  LENGTHS   OF   CURVES.  97 

The  evolute  of  the  c^'cloid  s  =  4a(l  —  cost)  is 
,^4^cKl-cosr) 


=  4asinT, 
cIt 

an  equal  cycloid,  with  its  vertex  at  the  origin. 


Examples. 

(1)  Prove  that  the  evolute  of  the  logarithmic  spiral  is  an 
equal  logarithmic  spiral. 

(2)  Find  the  evolute  of  a  parabola. 

(3)  Find  the  evolute  of  the  catenary. 

109.    The  evolute  of  an  epicycloid  is  a  similar  epicycloid,  with 
each  vertex  at  a  cusp  of  the  given  curve. 
Take  the  equation 


4b(a  +  b)  /^                 a        \        a  ^  -,/^.  r,-i 
5^ — ■ — ^    1  — cos -T   .     Art.  104ril. 


For  the  evolute. 


dl  1  —  cos  ■ 


4:b(a  +  b)      V  a  +  2b 


a  cIt 

Ab(a-{-b)   .        a  _^-, 

s  — !^ — ! — ^sm — ■ T.  ril 

a  +  2b         a  +  26  •-  -^ 

The  form  of  [1]  is  that  of  an  epicycloid  referred  to  a  vertex 
as  origin ;  let  us  find  a'  and  &',  the  radii  of  the  fixed  and  rolling 
circles. 

4  6'(a'  +  5')    .         a 

^^ — H — -  sui 

a'  a'-\-2b 

hence,  Ab'(a' +  b')  ^  4b  (a  +  b) 


s  =        -     .'   '  -sin    ,  r,       by  Art.  104  [2]  ; 

a +26' 

b  (a  +  b) 
a'  a  +  2  6    ' 


a'+2b'      a +  26 


98 


INTEGRAL   CALCFLUS. 


[Art.  110. 


Solving  these  equations,  we  get 


a'  = 


b'  = 


a  +  2b 

ah 
a +  26' 


a'  _  a 

h'~V 

and  the  radii  of  the  fixed  and  rolling  circles  have  the  same  ratio 
in  the  evolute  as  in  the  original  epicj'cloid ;  therefore  the  two 
curves  are  similar. 

Example. 

Show  that  the  evolute  of  a  hypocjxloid  is  a  similar  hypo- 
cycloid. 


110.  We  have  seen  that  in  involute  and  evolute  r  has  the  same 
value  ;  that  is,  t  =  t'. 

If  s'  and  t'  refer  to  the  evolute,  and  s  and  t  to  the  involute,  we 
have  found  that 

,     ds 


or 


di  ^ 

s'  = I,      I  being  a  constant, 

dr'  ^ 


the  length  of  the  radius  of  curvature  at  the  origin. 

{s'  +  l)dT'  =  ds, 
s=  C'(s'-tl)dT' 

Jo  . 

is  the  equation  of  the  involute. 

The  involute  of  the  catenary  s  =  a  tanr  is,  when  Z  =  0, 

s  =  a  I  tanT.cZr  =  a  log  sec  t,     the  tractrix. 


Chap.  IX.]  LEN^GTHS   OF   CUEVES.  99 

The  involute  of  tlie  cycloid  s  =  4  a  sinr  when  Z  =  0  is 

s  =  4a  I  sinr.fZr  =  4a(l  — cost), 

an  equal  cycloid  referred  to  its  cusp  as  origin. 

The  involute  of  a  C3'cloid  referred  to  its  cusp  s  =4a(l  —cost) 
when  Z  =  0  is 

s  =  4a  I  (1  — cos T)cZr  =  4a(T  +  sinr), 

a  curve  we  have  not  studied. 

The  involute  of  a  circle  s  =  clt  when  Z  =  0  is 


=  a  I  txIt  =  — 

Jo  i 


111.    While  any  given  curve  has  but  one  evolute,  it  has  an 
infinite  number  of  involutes,  since  the  equation  of  the  involute 


s'=  C\s-^l)dT 


contains  an  arbitrar}'  constant  I ;  and  the  nature  of  the  involute 
will  in  general  be  different  for  different  values  of  I. 

If  we  form  the  involute  of  a  given  curve,  taking  a  particular 
value  for  Z,  and  form  the  involute  of  this  involute,  taking  the  same 
value  of  Z,  and  so  on  indefinitel}',  the  curves  obtained  will  con- 
tinuall}'  approach  the  logarithmic  spiral. 

Let  s=/r  (1) 

be  the  given  curve. 

S  =jr'  {I  +fr)dT  =  It  +£fr.dT 
is  the  first  involute  ; 

is  the  second  involute  ; 

S=Zr+^  +  ^+ +^-^+   P'/t-^t"  (2) 

2      3!  n\     Jo  ^ 

is  the  wth  involute. 


X 


100  INTEGRAL    CALCULUS.  [Art.  112. 

B}^  Maclaurin's  Theorem, 

fr=f0  +  r/'o  +^/"0  +l!/'"0  + 

2  !  o  ! 

But  s  =  0  wlien  r  =  0  ;  hence  /o  =  0,  and 
2  !  3  ! 

-^  2  3!         4! 

»/o  3  !  4  !  0  ! 

0-^  (71+1)!       (w+2)!^(n  +  3)!  ^^ 

as  n  increases  indefinite!}''  all  the  terms  of  (3)  approach  zero 
(I.  Art.  133),  and  the  limiting  form  of  (2)  is 

Sz=It  -\ \- 

2!      3! 

=  ^A+I  +  Zl  +  Il  + _i\ 

Vl2!^3!  J 

s  =  Z(e^-  1)  by  I.  Art.  133  [2], 

which  is  a  logarithmic  spiral. 

112.    The  equation  of  a  curve  in  rectangular  coordinates  is 
readily  obtained  from  the  intrinsic  equation. 
Given  ,  s=/r, 

we  know  that  sin  r  =  — , 

as 

clx . 
and  cosT  =  -— J 

as 

hence  clx  =  costcZs  =  cost/'t.cZt, 

dy  =  sin  rds  =  sin  t/'txIt, 

COSrf'T.dr    I 
y  =  i  sin  t/'txIt 


Chap.  IX.]  LENGTHS   OF   CURVES.  101 

The  elimination  of  t  between  these  equations  will  give  us  the 
equation  of  the  curve  in  terms  of  a;  and  y.  Let  us  appl}-  this 
method  to  the  catenary. 

s  =  atanr, 

ds  =  a  sec^T. dr. 


C'        7          1         jl+sinr 
a;  =  a  I  secT.ar  =  a  log \ -. — , 

Jo  \1—  SUIT 

y  =  a  j  sec T tan T.dr  =  a(secT  —  1), 


—      1  +  sinr 


sinr  = 


1  —  sinr 

e<^  —1  _€«■  —  e  a 


secT  =  |-(e«-fe  «), 

2/  =  2  («"  +  «")  -  «' 
the  equation  of  the  catenary-  referred  to  its  lowest  point  as  origin. 


Curves  in  Space. 

113.  The  length  of  the  arc  of  a  curve  of  double  curvature  is 
the  limit  of  the  sum  of  the  chords  of  smaller  arcs  into  which  the 
given  arc  ma}'  be  broken  up,  as  the  number  of  these  smaller  arcs 
is  indefinitely  increased.  Let  (x,  y,  2) ,  (x-j-  dx,  y  -\-  Ay,  z  +  Az) 
be  the  coordinates  of  the  extremities  of  anj-  one  of  the  small  arcs 
in  question;  cLx,  Ay, Ae  are  infinitesimal ;  ^dx^-\-/\y--\- Az-  is  the 
length  of  the  chord  of  the  arc.  In  dealing  with  the  limit  of  the 
sum  of  these  chords,  scny  one  may  be  replaced  b}'  a  quantity  dif- 
fering from  it  b}^  infinitesimals  of  higher  order  than  the  first. 
VcZx"^  +  dy^-\-  dz^  is  such  a  value  ; 

1  VcZic^  +  dy^  +  dz^. 
I =10 


102 


INTEGRAL   CALCULUS. 


[Art.  113. 


Let  us  rectify  the  helix. 

x  =  a  cos  6  1 

y  =  asine    -.  (I.  Art. 214.) 

z  =  kO 
dx  =  —  asmd.cW, 
cly  =  a  cos 6. dO, 
dz  =  M6, 
ds-  =  (cr-{-Ji')de% 

s  =  {a'  +  k') h  C \l6  =  V^?TF {$1  -Bo). 


Examples. 

(1)  Find  the  length  of  the  curve  (y  =  -—,z  =  -^X 

Ans.    s=:x-{-z-\-L 

(2)  y  =  2'\/ax  —  x,z  =  x  —  %^—  Ans.   s  =  x  +  y  —  z. 


Chap.  X.] 


AREAS. 


103 


CHAPTER    X. 


AREAS. 


114.    "We  have  found  and  used  a  formula  for  the  area  bounded 
b}'  a  given  curve,  the  axis  of  X,  and  a  pair  of  ordinates. 


A 


-  I  ydx. 


Po-F 


We  can  readilj'  get  this  formula  as  a  definite  integral 
area  in  the  figiu'e  is  the  sum  of  the 
slices  into  which  it  is  divided  b}'  the 
ordinates  ;  if  Ax,  the  base  of  each 
slice,  is  indefinitely  decreased,  the 
slice  is  infinitesimal.  The  area  of 
any  slice  differs  from  yAx  by  less 
than  /1?/A.T,  which  is  of  the  second 
order  if  Ax  is  the  principal  infini- 
tesimal.    We  have  then 


The 


«o    *   Aa; 


^        limit  ="=^==1    . 


by  I.  Art.lGl 


Hence 


Xn 


Examples. 

1  xdy  is  the  area  bounded  by  a  curve,  the 

axis  of  Y,   and  perpendiculars  let  fall  from  the  ends  of  the 
bounding  arc  upon  the  axis  of  Y. 

(2)  If  the  axes  are  inclined  at  the  angle  w,  show  that  these 
formulas  become 


A.  =  siu  CO  I  ydx  =  sin  w  I  xdy. 


104 


INTEGRAL   CALCULUS. 


£Art.  ILo. 


115.    In  polar  coordinates  we  can  regard  the  area  between  two 
radii  vectores  and  the  curve  as  the  limit  of  the  sum  of  sectors. 

Tlie  area  in  question  is  the  sum 
of  the  smaller  sectorial  areas,  any 
one  of  whicli  differs  from  |-r^A^  b}- 
less  than  the  difference  between  the 
two  circular  sectors  ^(r  +  A?-)^A<^ 
and  ^?"A<^;   that  is,   by  less  tlian 

?-A?-A^  +  -^^ — ^ — ^,  which  is  of  the 

second  order  if  A^  is  the  principal  infinitesimal. 


Hence 


limit  \~,^.Z.^U..~\ 

^9. 


116.    Let  us  find  the  area  between  the  catenar}^  the  axis  of 
X,  the  axis  of  Y,  and  any  ordinate. 


but 
Hence 


Jx  Q   nx  X         _x 

yclx=  -  I  (e«  +  e  a)dx. 


A=-  (e«—  e  a) 

^  (e«  -  e'a)  =  s, 
A  =  as, 


by  Art.  90. 


and  the  area  in  question  is  the  length  of  the  arc  multiplied  b}-  tlie 
distance  of  the  lowest  point  of  the  curve  from  the  origin. 


117.    Let  us  find  the  area  between  the  tractrix  and  the  axis 
ofZ. 

We  have  dx  =  -^^  Va^  -  y"'.  (Art.  91.) 

y 


A=  i  ydx  =  —  j  dy\ld-  —  y'^. 


Chap.  X.]  ABEAS.  105 

The  area  in  question  is 

A  =  -Jdy  Va--  y'=^, 

which  is  the  area  of  the  quadrant  of  a  circle  with  a  as  radius. 

Example. 

Give,  by  the  aid  of  infinitesimals,  a  geometric  proof  of  the 
result  just  obtained  for  the  traetrix. 

118.  In  the  last  section  we  found  the  area  between  a  curve 
and  its  asymptote,  and  obtained  a  finite  result.  Of  course  this 
means  that,  as  our  second  bounding  ordinate  recedes  from  tlie 
origin,  the  area  in  question,  instead  of  increasing  indefinite^, 
approaches  a  finite  limit,  which  is  the  area  obtained.  Whether 
the  area  between  a  curve  and  its  asymptote  is  finite  or  infinite 
will  depend  upon  the  nature  of  the  curve. 

Let  us  find  the  area  between  an  hyperbola  and  its  asj^mptote. 

The  equation  of  the  h3-perbola  referred  to  its  as^-mptotes  as 


axes  IS  2 


xy  = 


a'  +  W 


Let  o)  be  the  angle  between  the  asymptotes  ;  then 

A  =  sin  CO  I  ydx  = ■ sm  w  I     —  =  oo  . 

Jo  4  Jo      X 

Take  the  curve          y~x  =  4a^(2a  —  x), 
or  2/'  =  4  a- ; 

X 

anj'  value  of  x  will  giA^e  two  values  of  y  equal  with  opposite 
signs ;  therefore  the  axis  of  x  is  an  axis  of  symmetry  of  the 
curve. 

When  x  =  2a^  y  —  0  ;  as  x  decreases,  y  increases  ;  and  when 
x  =  0,  y=  CO  .  If  a;  is  negative,  or  greater  than  2  a,  y  is  imagi- 
nary.    The  shape  of  the  curve  is  something  like  that  in  the 


106 


INTEGRAL    CALCULUS. 


[Art.  119. 


figure,  the  axis  of  F being  a,u  as3'mptote.    The  area  between  the 
curve  and  the  asymptote  is  then  either 

A=2  {  yclx     or    A=2  i  xdy  ; 

by  the  first  formula, 


A 


=*«fV 


2a  —  X 

X 


b}'^  the  second, 
A  =  Ua^ 

Examples. 


dy 


dx  —  4a^7r 


.  =  4a^7r. 


(1)  Find  the  area  between  the  curve  y'^{x^  +  cr)  =  a^x^  and  its 
asj'mptote  y  =  a.  Ans.    A  =  2a^. 

(2)  Find  the  area  between  y(2a  —  x)  =  x^  and  its  asj'mptote 
a;  =  2  a.  Ans.   A  =  3  tvoc. 

(3)  Find  the  area  bounded  b}'  the  curve  y'^= — - —       '  and 

n  'Y» 

its  as^'mptote  x^=a 

Ans.   A  =  2a'il-{-- 

119.  If  the  coordinates  of  the  points  of  a  curve  are  ex- 
pressed in  terms  of'an  auxiliary  variable,  no  new  difficulty  is 
presented. 

Take  the  case  of  the  circle  x^  +  y^  =  a^,  which  ma}^  be  written 

x  =  a  cos  </) 
y  =  a  sin  eft 
dy  =  acos(^dc}). 


The  whole  area    A  =  a^  I  cos^  cjidcf>  =  ttci^. 


Chap.  X.]  AREAS.  107 

Examples. 

(1)  The  whole  area  of  an  ellipse      ~  ,    .    ^  >•  is  irdb. 

y  =0  sill  cji  ) 

(2)  The  area  of  an  arch  of  the  cjxloicl  is  S-n-a^. 

(3)  The  area  of  an  arch  of  the  companion  to  the  cycloid 

X  —  a$,  y  z=a{\  —  cos6)  is  27ra^ 

120.  If  we  wish  to  find  the  area  between  two  curves,  or  the 
area  bounded  b}'  a  closed  curve,  the  altitude  of  our  elementary 
rectangle  is  the  difference  between  the  two  values  of  y,  which 
correspond  to  a  single  value  of  a).  If  the  area  between  two 
curves  is  required,  we  must  find  the  abscissas  of  their  points  of 
intersection,  and  the}'  will  be  our  limits  of  integration ;  if  the 
whole  area  bounded  by  a  closed  curve  is  required,  we  must  find 
the  values  of  x  belonging  to  the  points  of  contact  of  tangents 
parallel  to  the  axis  of  Y. 

Let  us  find  the  whole  area  of  the  curve 

49i7  9       4  O  7  0        0 

cry  +  b-x*  =  cro'x-, 

or  a*y''  =  b'-x\a^-x^). 

The  curve  is  s}Tnmetrical  with  reference  to  the  axis  of  X,  and 
passes  through  the  origin.  It  consists  of  two  loops  whose  areas 
must  be  found  separately.  Let  us  find  where  the  tangents  are 
parallel  to  the  axis  of  Y'. 

y=—x  V  a~  —  a^, 
a- 

dy      b    a^  —  2.^•^      , 

-^  =  -5 — z=^^  =  tanr. 

dx      a-  ^cr-x- 

T  —  -  when  tanr  =  00,  that  is,  when  x  =  ±  a. 

A  =  2-^  Cx  Va^  -  x\dx  +  2  ^  Cx  Va^  -  x'.dx  =  % ab. 
ccJ-a  a- Jo 


108  INTEGRAL   CALCULUS,  [Art.  121. 

Again  ;  find  the  whole  area  of  {y  ~xy  =  o?  —  y?. 

y  =  x  ±  Va^  —  x^, 

A  =  C(y'  -  y")dx  =  C2  Va-  -  x'. 
To  find  the  limits  of  integration,  we  must  see  where  t  =  -• 

dii      '\ld^  —  a?  ^  X  1  , 

-^  = —  =  00  when  cc  =  ±  a . 

dx         Va2  -  x^ 

A==2  r Va^  -x'  = 

tJ —a 


TTtt^ 


Examples. 

(1)  Find  the  area  of  the  loop  of  the  curve  y^  =  — ^^ — -^— — ^  • 

a  —  x 

Ans.    2ci'(l-- 

(2)  Find  the   area  between  the   curves  ?/"  — 4aa;  =  0   and 
^-^ay  =  0.  ^^,^^_    16a_^_ 

3 

(3)  Find  the  area  of  a  loop  of  a^y^  =  a:*(a^  —  x^) .     Ans.  — -. 

0 

(4)  Find  the  whole  area  of  the  curve 

2  y^  (a-  +  a;-)  -  4  ay  {a'  -x')  +  (a' -x^  =  0. 

A  "    (a       5V2' 

V  2 

121.    We  have  seen  that  in  polar  coordinates 

Let  us  tr}'  one  or  two  examples. 

(a)  To  find  the  whole  area  of  a  circle. 

The  polar  equation  is  r=a. 

A  =  l  \  c(?  d4>  =  7ra^. 


Chap.  X.]  AREAS.  109 

(b)  To  find  the  area  of  the  carclioide  r  =  2  a(l  —  cos  ^) . 

^4  =  1  (  4a-(l  -  co&cl,ydct>  =  2a-  i  (1-  2cos</)  +  cos- d4>)dcf>, 
A=Qa^7r. 

(c)  To  find  the  area  between  an  arch  of  the  epicycloid  and  the 
circumference  of  the  fixed  circle. 

a +  6 


x  =  (a  +  b)  cos  6  —  b  cos 

y  =  [a  +  b)sm6  —  b  sin  -^^  6 

We  can  get  the  area  bounded  by  two  radii  vectores  and  the 
arch  in  question,  and  subtract  the  area  of  the  corresponding 
sector  of  the  fixed  circle. 

Changing  to  polar  coordinates, 

X  =  r  cos  (f), 
y  =  r  sin  ^. 
^Ye  want  \  Crd^. 

y 

tan  d>  =  —i 

o  ,  7 ,      xdy—  ydx 
se&  (bdcb  =  — "^—^ —  ; 
ar 

but,  since  x  =  r  cos  <^,     sec  (^  =  -  ; 

X 

,                                       r^dcb      xdy— ydx 
hence  ~  =  — ^ — -J- — • , 

and  0'^  d(f)  =  xdy  —  ydx ; 

dx  =  (a-\-b)f-  sin  6  +  sin  ^^-t-^  q)  cW, 

dy  =  (a  +  b)  fcos 0  -  cos ^^-±-^ e\ dO. 

xdy  -  ydx  =  (a  +  b)  (a  +  2b)fl-  cos  -  oXw  =  7^dcf>. 

Our  limits  of  integration  are  obviously  0  and  — -■ 

a 


no 


INTEGRAL    CALCULUS. 

.25T 


[Art.  121. 


Hence  A  =  i(a  +  b){a-{-2b)J^   «  fl-cos'^6\cie, 

A  =  -{a  +  b){a  +  2b), 
a 

is  the  area  of  the  sector  of  the  epic3'cloicl.  Subtract  the  area  of 
the  cu'cular  sector  irab,  and  we  get 

j^^bUsi±2b)^ 
a 
as  the  area  in  question. 

(f?)  To  find  the  area  of  a  loop  of  the  curve  r^  =  a- cos  2  <^. 
For  any  value  of  cj)  the  values  of  r  are  equal  with  opposite 
signs.     Hence  the  origin  is  a  centre. 

When^  =  0,  r  =  ±a;   as  6  increases,  r  decreases  in  length 

tUl  ^  =  - ,  when  ?•  =  0  ;  as  soon  as  <^  >  -,  r  is  imagiuarv.     If  0 
4  4  " 

decreases  from  0,  r  decreases  in  length  until  0=  —  ^  when  r  =  0  ; 
and  when  ^<-,  r  is  imaginar3^  To  get  the  area  of  a  loop, 
then,  we  must  integrate  from  (^=  — jto  4>  =  y 

A  =  i  Cr\lcf>  =  \  a?  P  cos  2  c^.dtj  =  ^'• 


Examples. 


m 


(1)  Find  the  area  of  a  sector  of  the  parabola  r  =  — 
^   ^  1  +  cos  (^ 

(2)  Find  the  area  of  a  loop  of  the  curve  r-cos  (^  =  a- sin  3  4>. 

.       ■  3  a-      a- ,      -, 

Ans. log2. 

4        2     '' 

(3)  Find  the  whole  area  of  the  curve  r  =  a  (cos  2  (^  +  sin  2  <^) . 

Ans.    TTfr. 

(4)  Find  the  area  of  a  loop  of  the  curve  rcos^  =  a  cos  2  4>. 

Ans.    /'2-|yt-. 

(5)  Find  the  area  between  r  =a(sec<^+tan(^)  and  its  asymp- 


tote rcos^  =  2  a. 


Ans. 


|  +  2]«^. 


Chap.  X.] 


AREAS. 


Ill 


122.  When  the  equation  of  a  curve  is  given  in  rectangular 
coordinates,  we  can  often  simplify  the  problem  of  finding  its  area 
b}'  transforming  to  polar  coordinates. 

For  example,  let  us  find  the  area  of 

Transform  to  polar  coordinates. 

7-2  =  4  (a^  cos-  4>  +  b-  sin^  ^) , 
A=2i  {a- cos^ <li  +  b-sm^(f>) clef,  =  2 TV {a^  +  b'). 

Examples. 
(1)  Find  the  area  of  a  loop  of  the  curve  (a;-  +  y-y  =  4:  crx^y^. 

Ans.  — -• 


(2)  Find  the  whole  area  of  the  curve  1-  +  ^=—  1-  +  ^ 

a*      b*     c^Kcc 


Ans.  — —  (or  +  b") . 

(3)  Find  the  area  of  a  Ipop  of  the  curve  2/^  —  3  axy  +  a;^  =  0. 

Ans. 


123.    The  area  between  a  curve  and  its  evolute  can  easily  be 
found  from  the  intrinsic  equation  of  the  curve. 

It  is  easily  seen  that  the  area 
bounded  by  the  radii  of  curvature 
at  two  points  infinitely  near,  by 
the  curve  and  by  the  evolute,  dif- 
fers from  \f?dT  by  an  infinitesimal 
of  higher  order.  The  area  bounded 
b}'  two  given  radii  vectores,  the 
curve  and  the  evolute,  is  then 


A 


Hh 


112 


INTEGRAL   CA1,CULUS. 


[Art.  124. 


P  = 


ds 


Hence 


•'■^^Y*. 


ch 

For  example,  the  area  between  a  cycloid  and  its  evolute  is 
"i/d(4asinT) 


Let 


cIt 
rdr. 

To  =  0     and     t^ 


="''S 


^COS^rdr  =  2'7ra^. 


Examples. 

(1)  Find  the  area  between  a  circle  and  its  evolute. 

(2)  Find  the  area  between  the  circle  and  its  involute. 

Holditch's  Theorem. 

124.  If  a  line  of  fixed  length  move  with  its  ends  on  any  curve 
which  is  always  concave  toward  it,  the  area  between  the  curve 

and  the  locus  of  a  given  point 
of  the  moving  line  is  equal 
to  the  area  of  an  ellipse, 
of  which  the  segments  into 
which  the  line  is  divided  b}^ 
the  given  point  are  the  semi- 
axes. 

Let  the  figure  represent 
the  given  curve,  the  locus 
of  P,  and  the  envelope  of  the 
moving  line. 

Let^P=a  and  PB  =  h, 
and  let  CB  =  p,  C  being  the 
point  of  contact  of  the  moving  line  with  its  envelope.  Let 
AB  =  a-\-b  =  c. 


Chap.  X.J 


AREAS. 


113 


The  area  between  the  first  curve  and  the  second  is  the  area 
between  the  first  curve  and  the  envelope,  minus  the  area  between 
the  second  curve  and  the  envelope. 

Let  6  be  the  angle  which 
the  moving  line  makes  at 
any  instant  with  some  fixed 
direction.  Let  the  figure 
represent  two  near  positions 
of  the  moving  line  ;  A^,  the 
angle  between  these  posi- 
tions, being  the  principal  in- 
finitesimal. 

PB  =  p,  P'B'  =  p  +  Ap. 

The  area  PBB'P'P  differs 
from  ^p^dO  by  an  infinitesi- 
mal of  higher  order  than  the  first. 

ip^de  is  the  area  of  PBMP,  and  differs  from  PP'NB'  by  less 
than  the  rectangle  on  Plf  and  PQ,  which  is  of  higher  order  than 
the  first,  by  I.  Art.  L53.  But  PP'JSTB  diff"ers  from  PP'B'B  by 
less  than  the  rectangle  on  BN  and  NB',  which  is  of  higher  order 
than  the  first,  since  NB',  which  is  less  than  PP'+  Ap,  is  infini- 
tesimal and  A^  is  infinitesimal. 

The  area  between  the  first  curve  and  the  envelope  is  then 

^  I  p-d6 ;   or,  since  we  can  take  PP'A'A  just  as  well  for  our 
elementary  area,  ^  i  {c  —  pydd. 


^2-  ^2  7r 

Hence  -J  I  p-d$=^  I  (c  —  p)' 


dO 


whence 


j  pdd  = 


(1) 
The   area  between   the   second   curve  and  the   envelope   is 


i£(p  -  hfde. 


114 


INTEGRAL   CALCULUS. 


[Art.  125. 


The  area  between  the  first  curve  and  the  second  is  then 

,27r  ^2w 


=  bipd6-b^7r 


dd 


by  (1), 


=  '7rbc  —  b^Tr 

=  7rb(a  +  b)  —  b^TT, 
A  =  Tvab,  (2) 

which  is  the  area  of  an  eUipse  of  which  a  and  b  are  semi- axes. 


Q.  E.  D. 


Examples. 


(1)  If  a  line  of  fixed  length  move  with  its  extremities  on  two 
lines  at  right  angles  with  each  other,  the  area  of  the  locus  of  a 
given  point  of  the  line  is  that  of  an  ellipse  on  the  segments  of 
the  line  as  semi- axes. 

(2)  The  result  of  (1)  holds  even  when  the  fixed  lines  are  not 
perpendicular. 

Areas  by  Double  Integration. 

125.   If  we  choose  to  regard  x  and  y  as  independent  variables, 

we  can  find  the  area  bounded  by  two 
given  curves,  y  =  fx  and  y  =  Fx, 
by  a  double  integration.  Suppose 
the  area  in  question  divided  into 
slices  b}-  lines  drawn  parallel  to  the 
axis  of  Y,  and  these  slices  subdi- 
vided into  parallelograms  by  lines 
drawn  parallel  to  the  axis  of  X. 
The  area  of  any  one  of  the  small 
parallelograms  is  AyAx.  If  we 
keep  X  constant,  and  take  the  sum 
of  these  rectangles  from  y=fx  to  y  =  Fx,  we  shall  get  a  result 
differing  from  the  area  of  the  corresponding  slice  by  less  than 


Chap.  X.] 


AEEAS. 


115 


2  Aa;Ay,  which  is  infinitesimal  of  the  second  order  if  Ax  and  Ay 
are  of  the  first  order. 

Hence 


I  Ax.dy  =  Ax  idy 


is  the  area  of  the  slice  in  question.  If  now  we  take  the  limit  of 
the  sum  of  all  these  slices,  choosing  our  initial  and  final  values 
of  x^  so  that  we  shall  include  the  whole  area,  we  shall  get  the 
area  required. 

Hence  A=  (  1    ( dindx. 


=  f"(  f'«)' 

tJXg  \y/fX.  J 


In  writing  a  double  integral,  the  parentheses  are  usually  omit- 
ted for  the  sake  of  conciseness,  and  this  formula  is  given  as 


\      I  dydx, 


the  order  in  which  the  integrations  are  to  be  performed  being  the 
same  as  if  the  parentheses  were  actually  written. 

If  we  begin  b}-  keeping  y  constant,  and  integj-ating  with  respect 
to  cc,  we  shall  get  the  area  of  a  slice  formed  by  lines  parallel  to 
the  axis  of  X,  and  we  shall  have  to  take  the  limit  of  the  sum  of 
these  slices  varying  y  in  such  a  way  as  to  include  the  whole  area 
desu'ed.     In  that  case  we  should  use  the  formula 

j  dxdy. 

126.  For  example,  let  us  find  the  area  bounded  by  the  para- 
bolas 2/^  =  4  ax  and  a?  —  4:  ay. 

The  parabolas  intersect  at  the  origin  and  at  the  point  (4  a,  4  a). 


I     ^Jydx,   or  A=X    Xdxdy; 

ia 

dy  =  V  4  ax ; 


/y2 

4a 


ia 


I       I   diidx  =1        V4 ace ]dx=.  —  a^ 

Jo    J^  ^         Ja     \  4.a)  3 

4a 

The  second  formula  gives  the  same  result. 


116 


INTEGEAL   CALCTJLTJS. 


[Art.  127. 


Examples. 

(1)  Find  the  area  of  a  rectangle  by  double  integration  ;  of  a 
parallelogram  ;  of  a  triangle. 

(2)  Find  the  area  between  the  parabola  y^  =  ax  and  the  circle 


(3)  Find  the  whole  area  of  the  curve  {y  —  mx  —  cy  =  a^  —  x^ 

Ans.  TTtt^ 


127.    If  we  use  polar  coordinates  we  can  still  find  our  areas 
by  double  integration. 

Let  r  =f(f>  and  r  =  Fcfi 
be  two  curves.  Divide  the 
area  between  them  into 
slices  by  drawing  radii 
vectores ;  then  subdivide 
these  slices  by  drawing 
arcs  of  circles,  with  the 
origin  as  centre. 

Let  P,  with  coordinates 
r  and  ^,  be  any  point 
within  the  space  whose 
area  is  sought.  The  curvilinear  rectangle  at  P  has  the  base  rA<^ 
and  the  altitude  Ar  ;  its  area  differs  from  r/\(jiAr  by  an  infinitesi- 
mal of  higher  order  than  rA<^A?\ 

The  area  of  any  slice  as  aba'b'  is   I  rAcfiClr,  (f}  and  Acfi  being 

constant,  that  is  A<^  |  rdr.      The  whole  area,  the  limit  of  the 

sum  of  such  slices  is  ^=  I      I  rdrd^.  (1) 

Or  we  may  first  sum  our  rectangles,  keeping  r  unchanged, 
and  we  get  as  the  area  of  efe'f 

rAr  I    d<j>^    and   -4=1      I  rdcf>dr.  (2) 


'i-H 


</- 


Chap.  X.] 


AEEAS. 


117 


For  example,  the  area  between  two  concentric  circles,  r  =  a 
and  r  =  6,  is 

A—  \     I  rd(j)dr  =  |       |  rdrdcji  =  7r(a^  —  b^) . 

Again,  let  us  find  the  area  between  two 
tangent  circles  and  a  diameter  through  the 
point  of  contact. 

Let  a  and  b  be  the  two  radii, 

?*=2acos<^  (1) 

and  ?'  =  2  &  cos  <^  (2) 

are  the  equations  of  the  two  circles. 

j  rdrd^  =  2  (a^  -  6^)  (  cos^  ^dcf^  =  ^  (a^  -  6^) . 

0    »/2  6  cos  (p  c/0  2 

If  we  wish  to  reverse  the  order  of  our  integrations  we  must 
break  our  area  into  two  parts  by  an  arc  described  from  the  origin 
as  a  centre,  and  with  2  &  as  a  radius  ;  then  we  have 

2b      cos-i^  2  a      cos-i|^ 

^  =  I       I  rdcjido'  +  I        I  rd<f)dr 
Jo     J       r  Jib  Jo 

COS-I26 

J^26/              .J.  y  \  y^2a  ^. 

I   ricos"^^^ cos~^r— ]d?'-f-  I  7*cos~^T— 'dr 
0     V         2a  ^b)       '  Jib  2a 

2^  ^ 


Example. 

Find  the  area  between  the  axis  of  X  and  two  coils  of  the 
spiral  r=  a(f>. 


118 


INTEGRAL    CALCULUS. 


[Art.  128. 


CHAPTER    XI, 


AREAS    OF    SUKFACES. 


Surfaces  of  Revolution. 

128.  If  a  plane  curve  y  =fx  revolves  about  the  axis  of  X,  the 
area  of  the  surface  generated  is  the  limit  of  the  sum  of  the  areas 
generated  by  the  chords  of  the  infinitesimal 
arcs  into  which  the  whole  arc  may  be  broken 
up.  Each  of  these  chords  will  generate  the 
surface  of  the  frustum  of  a  cone  of  revolu- 
tion if  it  revolves  completely  around  the 
axis  ;  and  the  area  of  the  sm^face  of  a  frus- 
tum of  a  cone  of  revolution  is,  by  element- 
ar}'  Geometry,  one-half  the  sum  of  the  cir- 
cumferences of  the  bases  multiplied  by  the  slant  height.  The 
frustum  generated  by  the  chord  in  the  figure  will  have  an  area 
differing  by  infinitesimals  of  higher  order  from  ir{y  -\-y  +  Ay)  As 
or  from  27ryds.     The  area  generated  by  any  given  arc  is  then 


Ax 


s 


=  2  77  I  yds 


[1] 


If  the  arc  revolVes  through  an  angle  B  instead  of  making  a 
complete  revolution,  the  surface  generated  is 


S^B 


»  \yds. 


[2] 


Example. 
Show  that  if  the  arc  revolves  about  the  axis  of  y, 


S 


=  27r  I  xds. 


Chap.  XI.] 


AREAS   OF   SUEFACES. 


119 


129.    To  find  the  area  of  a  cj^linder  of  revolution. 

Take  the  axis  of  the  cylinder  as  the  axis  of  X.     Let  a  be  the 
altitude  and  b  the  radius  of  the  base  of  the 
cylinder.      The  equation  of  the  revolving 
line  is 

2/  =  6; 

ds  =  VcZcc^  +  dy^  =  dx ; 
S=27r  i  ydx  =  2 TTCib, 

or  the  product  of  the  altitude  by  the  circumference  of  the  base. 
Again,  let  us  find  the  surface  of  a  zone. 
The  equation  of  the  generating  circle  is 

af  -\-y^  =  a^ ; 

adx 


ds  = 


y 


X(j      x^ 


S  =  2'7r  i  adx  =2 aiT(xi  —  Xo). 
If  Xq  =  —  a  and  ccj  =  a, 

Hence  the  surface  of  a  zone  is  the  altitude  of  the  zone  multi- 
plied by  the  circumference  of  a  great  circle,  and  the  surface  of  a 
sphere  is  equal  to  the  areas  of  four  great  circles. 

Again,  take  the  surface  generated  by  the  revolution  of  a 
cycloid  about  its  base. 


x=:a6  —  a  sin  i 
y  =  a  —  a  cos  $ 


V 


ds  =  adO  V2  ( 1  -  cos  6) ,  by  Art.  94  ; 

aS  =  2 TT  raV2 .  (1  -  cos 0)i  dd  =  ^iro?. 


120  IKTEGEAL   CALCULUS.  [Art.  130. 

Examples. 
(1)  The  area  of  the  surface  generated  by  the  revolution  of  the 
ellipse  ^4-^=1 

o?    y" 

about  the  axis  of  X  is  2  ircib  J  Vl  —  e^  +  ^j ; 

about  the  axis  of  Fis  27raM  1  +    ~  ^  log  —^\ 

V  2  e  1—eJ 

where  e^= — . 


(2)  Find  the  area  of  the  surface  generated  by  the  revolution 
of  the  catenarj'  about  the  axis  of  X ;  about  the  axis  of  Y. 

(3)  The  whole  surface  generated  by  the  revolution  of  the 
tractrix  about  its  asj'mptote  is  47ra^. 

(4)  The  area  generated  b}'  the  revolution  of  a  cj'cloid  about 
its  vertical  axis  is  8  -n-a^^Tr  —  |) . 

(5)  The  area  generated  b}'  the  revolution  of  a  cycloid  about 
the  tkngent  at  its  vertex  is  ^^--n-a^. 

130.  If  we  know  the  area  generated  by  the  revolution  of  a 
curve  about  an}'  axis,  we  can  get  the  area  generated  b}-  the 
revolution  about  any  parallel  axis  by  an  easy  transformation  of 
coordinates. 

Given  the  surface  generated  by  the  arc  from  Sq  to  Si  about 

OX,  to  find  the  area  generated  by 
the    same    arc   when   it    revolves 
-x'  about  OX'. 

Let  S  be  the  surface  about  OX, 
and  S'  about  O'X'. 
We  have 


Po 


/S  =  2  TT  Cycls,     S'=  2  TT  Cy'ds'. 


Chap.  XI.]  AREAS  OF  SURFACES.  121 

By  Anal.  Geom.,  x  =  x', 

y  =  yo  +  y'- 

Hence  clx  =  dx',     dy  ==  dy',     els  =  ds', 

and        /S  =  2 TT  C(yo  +  y')ds  =  2 7ri/o(si  -  Sq)  +  2-nr  Cy^'ds, 

=  27ryo{Si  —  So)-\-S'. 

Therefore  ^'  =  ^S  _  2  7r?/o(si  -  Sq)  .  [1] 

Si  —  So  is  the  length  of  the  revolving  curve ;  2  ttt/o  is  the  cir- 
cumference of  a  circle  of  which  y^  is  the  radius.  Hence  the  new 
area  is  equal  to  the  old  area  minus  the  area  of  a  cj'linder  whose 
length  is  the  length  of  the  given  arc  and  whose  base  is  a  circle 
of  which  the  distance  between  the  two  lines  is  radius. 

In  using  this  principle  careful  attention  must  be  paid  to  the 
sign  of  i/o,   and  it  must  be  noted  that  the   original   formula 


S  = 


=  2  TT  j  yds  will  alwa3's  give  a  negative  value  for  the  area  of 

the  surface  generated,  if  the  revolving  arc  starts  from  below  the 
axis  ;  and  hence,  that  the  surface  generated 
by  the  revolution  of  any  curve  about  an 
axis  of  sj'mmetry  will  come  out  zero. 

As  an  example  of  the  use  of  the  princi- 
ple, let  us  find  the  surface  of  a  ring. 

Let  a  be  the  distance  of  the  centre  of  — 
the  ring  from  the  axis,  and  b  the  radius  of 
the  circle.     Since  the  area  generated  by  the 
revolution  of  the  circle  about  a  diameter  is  zero,  the  required 
area  is 

2Trb.27ra  =  4:Trhib. 

Example. 

Find  the  area  of  the  ring  generated  by  the  revolution  of  a 
cycloid  about  any  axis  parallel  to  its  base. 


Ans.    S  =  AabTr[Tr-\ ■ ). 


122  INTEGRAL   CALCULUS.  [Art.  131. 

131.    If  we  use  polar  coordinates, 

jS  =27r  I  yds 

becomes  S  =  2 it  \  r sm^.ds. 


where  ds  =  Vc?r^  +  tM^^. 

For  example  ;  let  us  find  the  area  of  the  surface  generated  by 
the  revolution  of  the  upper  half  of  a  cardioide  about  the  hori- 
zontal axis. 

?•=  2a(l—  cos^)  ; 

dr  =  2asin<^.d<^, 

ds^  =  8  a^  ( 1  —  cos  <^)  d<^^, 

^  =  27r  j  4  V2a2(l- cos<^)'siB«^.d(^. 

EXAIVIPLES. 

(1)  Find  the  surface  of  a  sphere  from  the  polar  equation. 

(2)  Find  the  surface  of  a  paraboloid  of  revolution  from  the 
polar  equation  of  the  parabola 

m 


r  = 


1  —  cos  ^ 

Any  Surface. 

132.  Let  X,  y,  z  be  the  coordinates  of  an}^  point  P  of  the  sur- 
face, and  X  +  Ax,  y  +  Ay,  z-\-  Az  the  coordinates  of  a  second 
point  Q  infinitely  near  the  first.  Draw  planes  through  P  and  Q 
parallel  to  the  planes  of  XY"  and  YZ.  These  planes  will  inter- 
cept a  curved  quadrilateral  PQ  on  the  sm-face  ;  its  projection  pg, 
a  rectangle,  on  the  plane  of  XZ\  and  a  parallelogram  p'q'  not 
shown  in  the  figure  on  the  tangent  plane  at  P,  of  which  pq  is 


Chap.  XI.] 


AREAS   OF   SUEFACES. 


123 


the  projection.  PQ  will  differ  from  p'q'  by  an  infinitesimal  of 
higher  order,  and  therefore  our  required  surface  will  be  the  limit 
of  the  sum  of  the  pai'allelograms  of  which  p'q'  is  any  one. 


If  (3  is  the  angle  the  tangent  plane  at  P  makes  with  XZ, 
2y'q' cos  ft  =  2)q  or  p'q' =2)q  sec  ft.  =  Ax  Az  sec  (3,  and  o-,  our  sur- 
face required,  is  equal  to  the  double  integral  o-  =  |  |  sec  ftdxd2 
taken  between  limits  so  chosen  as  to  embrace  the  whole  surface. 

The  equation  of  the  tangent  plane  is 

{x-x,)D^J+{y-y,)DyJ+{z-z,)D,J=0,   by  I.  Art.  217, 

(a;oi2/oi^o)  standing  for  the  coordinates  of  the  point  of  contact, 

and  f{x.,y,z)  =  0  being  the  equation  of  the  surface. 

The  direction  cosines  of  the  perpendicular  from  the  origin  upon 

the  plane  are 

D.J 

cos  a  =     ,  „  :? 

DyJ 

'^{D.jy  +  iDyjY  +  iD.jY 

DzJ 

COS  v  ^=  —  -   ■  5 

^     '^{D.Jf  +  iDyjy  +  iD.JY 
by  Anal.  Geom.  of  Three  Dimensions. 


Q,osft= 


124  INTEGRAL   CALCULUS.  [Art.  132. 

Hence,  dropping  the  accents, 


By  considering  the  projections  upon  the  other  coordinate  planes 
we  shall  find 

,^^j^(Djy-  +  WY+iI>.f)\yay_  1-3] 

In  each  of  the  formulas  the  derivatives  are  partial  derivatives. 
Let  us  find  the  area  of  the  portion  of  the  surface  of  the  sphere 

a?  -\- 'if  -\- z^  =  Cf? 

intercepted  by  the  three  coordinate  planes. 

^{Djy  +  {Djy+  (Djy  =  2a.  • 

o-=  I     I  -dydz;  (1) 

Jo  Jox 

or  o-=  f  \-dzdx;  (2) 

Jo  Joy 

or  o-=  \     \-dxdy.  (3) 

Jo  Joz 

For,  in  the  second  one,  which  agrees  best  with  the  figure,  we 
must  take  our  limits  so  that  the  limit  of  the  sum  of  the  projec- 
tions may  be  the  quadi-ant  in  which  the  sphere  is  cut  by  the 


Chap.  XI.] 


AEEAS   OF    SURFACES. 


125 


plane  XZ ;  and  the  equation  of  this  section  is  obtained  hj  letting 
2/  =  0  in  the  equation  of  the  sphere,  and  is 

whence  z  =  Va^  —  af. 

If  we  take  as  our  limits  in  the  integral  (  -  clz  zero  and  -s/ar—  x^ 

^  y 

we  shall  get  the  area  whose  projection  is  a  strip  running  from 
the  axis  of  Z  to  the  curve  ;  then,  taking  j  (  j  -  f^ )  f?^  from  0  to 

a,  we  shall  get  the  area  whose  projection  is  the  sum  of  all  these 
strips,  and  that  is  om-  required  surface. 


y 


=  Va^_a;2. 


(T=al     I  — 

Jo  Jo  Va^  -x'-z^ 


f 


clz 


Vci^  —  3Cp  — 


=  sin"-^  - 


Vcr  —  af 


if  we  regard  x  as  constant ; 


c7o- 


>/rt2-a;2 


dz 


cr  —  a  \     -ax  =  — , 

Jo    2  2 

the  required  area.     Formulas  (1)  and  (3)  give  the  same  result. 

133.  Suppose  two  cylinders  of  revolution  drawn  tangent  to 
each  other,  and  perpendicular  to  the  plane  of  a  great  circle  of  a 
sphere,  each  having  the  radius  of  the 
great  circle  as  a  diameter ;  required  the 
surface  of  the  sphere  not  included  by 
the  cylinders. 

The  surface  required  is  eight  times 
the  surface  of  which  the  shaded  portion 
of  the  figure  is  the  projection. 

If  we  take  the  plane  of  the  great 
cu'cle  as  the  plane  of  XY, 


126  INTEGRAL   CALCULUS. 

Q(?  —  ax  +  y-  =  (d 
is  the  equation  of  the  cylinder,  and 

o(?  -\- 'if  -\- z^  =  a? 


[Art.  133. 
(1) 


of  the  sphere. 

We  have  o-  =  I    1  ■ 
From  (2) 


(2) 


DJ=2x, 

{Djy+{Djy+{j)jy  =  ia\ 

dydx 


dydx. 


/»  ra  /•  /»        ayax 

Hence        c.=J  j  J ^^^^- =' «J  J  v^^=^ 


r 


Our  limits  of  integration  for  y  are  Vaa;  —  oi?  and  Va^  — a^;  for 
X  are  0  and  a. 

-^a-^-x-^clydx 

'^ax-x'^ 


J  Vtt"  —  ce^  — 


—  ain     1 


2/^ 


Va^- 


= Sin" 

9 


'^l: 


a  +  ^ 


To  find  j  sin-^-\U^ — dx  we  must  integrate  by  parts. 
Jo  \  a  +  £c 


Let 
and 


.   _i    I     re 

w  =  sm  ^A  , 

\  a  +  a; 

dv  =  dx' ; 

du  = --1  -.(Zic ; 

2(a4-x)  \x 


Jsin-^. 


a  +  cc 


,dx  =  a  sin 


\a-\-x        "i  J  a+x 


Chap.  XI.]  AREAS  OF   SUEFACES.  127 

Let  to=  -^x;     2  wcho  —  dx 

,„a        C:J^  =  2  f!^,  =  2  ff  1  -  -^)aw. 

'^"^^  J  a  +  x         J  a  +  w"        J\       a-\-io'J 

/'Jx.dx     „/  ,    ,     _i  w" 

-^ =  2[io  —  'Ja tan  ^  — = 
a  +  x        \         ^  ^a 

Jsin'-^-y dx 
0           \a-\-x  I- 

=  asin"^-^  +  «tan-U  —  a  =  —  +  —  —  a  =  —  —a, 
2  4        4  2 

V2     2     ; 

8cr  =  So?  is  the  whole  surface  in  question. 

Examples. 

(1)  Find  the  area  included  by  the  cylinders  described  in  Art. 
133  b}'  direct  integration. 

(2)  A  square  hole  is  cut  through  a  sphere,  the  axis  of  the 
hole  coinciding  with  a  diameter  of  the  sphere  ;  find  the  area  of 
the  surface  removed. 

(3)  A  cj'linder  is  constructed  on  a  single  loop  of  the  curve 
r=acos?i^,  having  its  generating  lines  perpendicular  to  the 
plane  of  this  cm've ;  determine  the  area  of  the  portion  of  the 
surface  of  the  sphere  a?  +y--\-z'  =  a-  which  the  cylinder  inter- 

«ept«-  Ans.   ^(|-l). 

(4)  The  centre  of  a  regular  hexagon  moves  along  a  diameter 
of  a  given  circle  (radius  =  a) ,  the  plane  of  the  hexagon  being 
perpendicular  to  this  diameter,  and  its  magnitude  var3ing  in 
such  a  manner  that  one  of  its  diagonals  alwaj's  coincides  with 
a  chord  of  the  cii'cle ;   find  the  surface  generated. 

Ans.   a2(27r  +  3V3). 


128  INTEGRAL  CALCULUS.         [Art.  134. 


CHAPTER    XII. 

VOLUMES. 

Single  Integration. 

134.  If  sections  of  a  solid  are  made  b}'  parallel  planes,  and  a 
set  of  cylinders  drawn,  each  having  for  its  base  one  of  the  sec- 
tions, and  for  its  altitude  the  distance  between  two  adjacent 
cutting  planes,  the  limit  of  the  sum  of  the  volumes  of  these 
cj'linders,  as  the  distance  between  the  sections  is  indefinitely 
decreased,  is  the  volume  of  the  solid. 

We  shall  take  as  established  b}'  Geometrj^  the  fact  that  the 
volume  of  a  C3'linder  or  prism  is  the  product  of  the  area  of  its 
base  b}'  its  altitude. 

It  follows  from  what  has  just  been  said,  that  if,  in  a  given 
solid,  all  of  a  set  of  parallel  sections  are  equal,  the  volume  of 
the  solid  is  its  base  hy  its  altitude,  no  matter  how  irregular  its 
form. 

Let  us  find  the  volume  of  a  pjTamid  having  b 
J\         for  the  area  of  its  base,  and  a  for  its  altitude. 
//  '\  Divide  the  pyramid  b}'  planes  parallel  to  the 

//     'A       base,  and  let  z  be  the  area  of  a  section  at  the  dis- 
/- -/ — i\      tance  x  from  the  vertex.  , 

/  y H         We  linow  from  Geometry  that  -  =  - . 

/     / \  \  h      cc 

\  I             ■■•A        TT  &    " 
V               'A        Hence                      ^  =  -<>  ^- 
'                                              a- 

Let  the  distance  between  two  adjacent  sections  be  dx ;  then 

the  volume  of  the  cylinder  on  z  is 

and  F,  the  required  volume  of  the  pyramid,  is 
V=—  \  x-dx  =  — . 


Chap.  XII.]  VOLUMES.  129 

Precisel}'  the  same  reasoning  applies  to  any  cone,  which  will 
therefore  have  for  its  volume  one-third  the  product  of  its  base 
by  its  altitude. 

Example. 

Find  the  volume  of  the  frustum  of  a  pyramid  or  of  a  cone. 

135.  If  a  line  pKRce^ keeping  always  parallel  to  a  given  plane, 
and  touching  a  plane  curve  and  a  straight  line  parallel  to  the 
plane  of  the  curve,  the  surface  generated  is  called  a  conoid. 
Let  us  find  the  volume  of  a  conoid  when  the  director  line  and 
curve  are  perpendicular  to  the  given  plane. 

Divide  the  conoid  into  laminae  by 
planes  parallel  to  the  fixed  plane. 

Let  Ay  be  the  distance  between 
two  adjacent  sections,  and  let  x  be 
the  length  of  the  line  in  which  any 
section  cuts  the  base  of  the  conoid  ; 
let  o  be  the  altitude  and  h  the  area 
of  the  base  of  the  figure.  Any  one  of  our  elementary  C3"linders 
will  have  for  its  volume  |^aa;A?/,  since  the  area  of  its  triangular 

base  is  ^ax,  and  we  have  V—^a  |  xdy.,  the  limits  of  integra- 
tion being  so  taken  as  to  embrace  the  whole  solid.  |  xdy  be- 
tween the  limits  in  question  is  the  area  of  the  base  of  the  CO" 
noid  ;  hence  its  volume, 

EXABIPLES. 

(1)  Find  the  volume  of  a  conoid  when  the  director  line  and 
curve  are  not  perpendicular  to  the  given  plane. 

(2)  A  woodman  fells  a  tree  2  feet  in  diameter,  cutting  half- 
way through  from  each  side.  The  lower  face  of  each  cut  is 
horizontal,  and  the  upper  face  makes  an  angle  of  45°  with  the 
horizontal.     How  much  wood  does  he  cut  out? 


130  INTEGEAI.   CALCULUS.  [Art.  136. 


136.    To  find  the  volume  of  an  ellipsoid. 

(J?      h^      c^ 


Take  the  cutting  planes  parallel  to  the  plane  of  XY".     A  sec- 
tion at  the  distance  z  from  the  origin  will  have 

c^     y^ _.  _^'_  c^  —  z^ 

a   I h    I 

for  its  equation,  and  —  Vc^  —  z"^  and  -  Vc^  —  z^  for  its  semi-axes  ; 

hence  its  area  will  be  — —  (c^  —  z') . 
& 

Any  of  the  elementary  cylinders  will  have  for  its  volume 
^^(c^  — 2;^)A2:,  and  we  shall  have  for  the  whole  solid 
■KCib 


If  a,  &,  and  c  are  equal,  the  ellipsoid  is  a  sphere,  and 

Examples. 

(1)  Find  the  volume  included  between  an  hyperboloid  of  one 
sheet 

and  its  asymptotic  cone 

^  _l_^ ?-  =  0. 

G?        IP'        (^ 

Ans.  It  is  equal  to  a  cylinder  of  the  same  altitude  as  the 
solid  in  question,  and  having  for  a  base  the  section  made  by  the 
plane  of  XY". 

(2)  Find  the  whole  volume  of  the  solid  bounded  by  the  surface 

^j^tj^t  =  \.  STTCtbc 

r,2  ^  7,2  ^  p4  Ans.   — - — 


Chap.  XII.]  VOLUMES.  131 

(3)  Find  the  volume  cut  from  the  surface 

c       b 

by  a  plane  parallel  to  the  plane  of  (  TZ)  at  a  distance  a  from  it. 

Ans.    7r«^^(6c), 

(4)  Find  the  whole  volume  of  the  solid  bounded  by 

(ay^ -{- y^  +  z-y  =  27 a^xtjz.        ■       A7is.    fa^ 

(5)  The  centre  of  a  regular  hexagon  moves  along  a  diameter 
of  a  given  circle  (i-adius  =  a),  the  plane  of  the  hexagon  being 
perpendicular  to  this  diameter,  and  its  magnitude  varying  in 
such  a  manner  that  one  of  its  diagonals  alwaj's  coincides  with 
a  chord  of  the  circle ;   find  the  volume  generated. 

Ans.    2y3.a^. 

(6)  A  circle  (radius  =  a)  moves  with  its  centre  on  the  cir- 
cumference of  an  equal  circle,  and  keeps  parallel  to  a  given 
plane  which  is  perpendicular  to  the  plane  of  the  given  circle  ; 
find  the  volume  of  the  solid  it  will  generate,  9   3 

A71S.    f^(37r  +  8). 
o 

Solids  of  Revolution.     Single  Integration. 

137.  If  a  solid  is  generated  b}*  the  revolution  of  a  plane  curve 
y  =  fx  about  the  axis  of  x,-  sections  made  by  planes  perpendicu- 
lar to  the  axis  are  circles.  The  area  of  any  such  circle  is  Try-, 
the  volume  of  the  elementar}"  cylinder  is  iry^Ax,  and  ' 


V  = 


/^  X 

r  I   y^dx 


is  the  volume  of  the  solid  generated. 

For  example  ;  let  us  find  the  volume  of  the  solid  generated  by 
the  revolution  of  one  branch  of  the  tractrix  about  the  axis  of  X. 
Here  we  must  integrate  from  a;  =  0  to  a;  =  00 . 


V 


y^dx. 


132  INTEGRAL   CALCULUS.  [Art.  138. 

We  have  dx  =  -  ^"^  ~  y''>" dy  Art.  91  [2] 

y 

in  the  case  of  the  traetrix  ; 

hence  V=  ~  tc  \  y{a^  —  y^Y^dy. 

When  a;  =  0,  y  =  a^  and  when  re  =  oo,  y  =  0. 
Therefore  F=  — tt  I  y{a^  —  y^)idy  = 


ttCV 


Examples. 

(1)  Kthe  plane  curve  revolves  about  the  axis  of  Y, 

V—TT  I    x-dy. 

(2)  The  volume  of  a  sphere  is  %  -n-ce. 

(3)  The  volume  of  the  solid  formed  b}"  the  revolution  of  a 
C3'cloid  about  its  base  is  5  tt^  cc^. 

(4)  The  curve  y^{2a  —  x)  =  x^  revolves  about  its  asymptote  ; 
show  that  the  volume  generated  is  2TT^a^. 

Solids  of  Revolution.     Double  Integration. 

138.  If  we  suppose  the  area  of  the  revolving  curve  broken  up 
into  infinitesimal  rectangles  as  in  Art.  125,  the  element  AxAy 
at  any  point  P,  whose  coordinates  are  x  and  y,  will  generate 
a  ring  the  volume  of  which  will  differ  from  27ry^xAy  by  an 
amount  which  will  be  an  infinitesimal  of  higher  order  tlian  the 
second  if  we  regard  Ax  and  Ay  as  of  the  first  order.  For 
the  ring  in  question  is  obviousl}^  greater  than  a  prism  having 
the  same  cross-section  AxAy,  and  ha-ving  an  altitude  equal  to  the 
inner  circumference  2  Try  of  the  ring,  and  is  less  than  a  prism 
having  Ax  Ay  for  its  base  and  27r{y  -j-  Ay),  the  outer  circumfer- 
ence of  the  ring,  for  its  altitude  ;  but  these  two  prisms  difier  by 
2irAx(Ayy,  which  is  of  the  third  order. 


Chap.  XII.]  VOLUMES.  133 

Aa;  I  2  irydy,  where  the  upper  limit  of  integration  is  the  ordi- 
nate of  the  point  of  the  curve  immediately  ahove  P,  and  must  be 
expressed  in  terms  of  x  hy  the  aid  of  the  equation  of  the  revolv- 
ing curve,  will  give  us  the  elementary  cylinder  used  in  Art.  137. 

The  whole  volume  required  will  be  the  limit  of  the  sum  of 
these  c^iinders  ;  that  is, 

V=2-^r  fydydx.  [1] 

If  the  figure  revolved  is  bounded  by  two  curves,  the  required 
volume  can  be  found  by  the  formula  just  obtained,  if  the  limits 
of  integration  are  suitably'  chosen. 

Let  us  consider  the  following  example  : 

A  paraboloid  of  revolution  has  its  axis  coincident  with  the 
diameter  of  a  sphere,  and  its  ^-ertex  in  the  surface  of  the  sphere  ; 
required  the  volume  between  the  two  surfaces. 

Let  y^=2  7nx  (1) 

be  the  parabola,  and     of  -\-y^  —  2  ax  =  0  -  (2) 

be  the  circle,  which  form  the  paraboloid  and  the  sphere  by  then- 
devolution.  The  abscissas  of  their  points  of  intersection  are  0 
and  2  (a  —  7n) . 

We  have  V=2Tri   i  ydydx, 

and,  in  performing  our  first  integration,  our  limits  must  be  the 
values  of  y  obtained  from  equations  (1)  and  (2). 

We  get  F=  TT  j  [2 (a  —  m)x  —  xr']dx, 

and  here  our  Imiits  of  integration  are  0  and  2  (a  —  m). 

Hence  F=  |7r(a  —  7?i)^  = -— , 

6 

if  7i  is  the  altitude  of  the  solid  in  question. 

Examples. 

(1)  A  cone  of  revolution  and  a  paraboloid  of  revolution  have 
the  same  vertex  and  the  same  base ;  required  the  volume  be- 
tween  them.        ^^^_   ^rrf  ^^^^^  j^  .^  ^^^  ^^^.^^^^  ^^  ^^^  ^^^^^_ 


134  INTEGRAL   CALCULUS,  [Art.  139. 

(2)  Find  the  volume  included  between  a  right  cone,  whose 
vertical  angle  is  60°,  and  a  sphere  of  given  radius  touching  it 
along  a  circle.  j^^^^    ■^ 

6  ■ 

Solids  of  Revolution.     Polar  Formula. 

139.  If  we  use  polar  coordinates,  and  suppose  the  revolving 
area  broken  up,  as  in  Art.  127,  into  elements  of  which  rclt^dr 
is  the  one  at  any  point  P  whose  coordinates  are  r  and  <^,  the 
element  rdcftdr  will  generate  a  ring  whose  volume  will  differ 
from  2777-^  sin  (/)d(/)C?r  by  an  infinitesimal  of  higher  order  than  the 
second,  if  we  regard  dcfi  and  d?'  as  of  the  first  order ;  for  it  will 
be  less  than  a  prism  having  for  its  base  rdcfidr,  and  for  its  alti- 
tude 2  IT (r  +  dr)  sin {(f>-\-d(p),  and  greater  than  a  prism  having 
the  same  base  and  the  altitude  2  Trr  sin  (ft ;  and  these  prisms 
differ  b}'  an  amount  which  is  infinitesimal  of  higher  order  than 
the  second. 

"We  shall  have  then 

F=  2  TT  rp-^  sin  c^cZrdc^,  [1] 

the  limits  being  so  taken  as  to  bring  in  the  whole  of  the  gener- 
ating area. 

For  example  ;  let  us  find  the  volume  generated  by  the  revolu- 
tion of  a  cardioide  about  its  axis. 

?'=  2a(l  —  cos^) 
is  the  equation  of  the  cardioide  ; 


'  F=  2  TT  ffr^ sin <^dr#. 


Our  first  integral  must  be  taken  between  the  limits  r  =  0  and 
r  =  2  a  (1  —  cos  0) ,  and  is 

(1—  cos^)®sin<^d^. 

3 


16      r'^ 

'= — a^TT  I  (1— cos^)^sin(^(^d), 
3        Jo 


3 


Chap.  XII.] 


VOLUMES. 


135 


Example. 

A  right  cone  has  its  vertex  on  the  surface  of  a  sphere,  and  its 
axis  coincident  with  the  diameter  of  the  sphere  passing  through 
that  point ;  find  the  volxune  common  to  the  cone  and  the  sphere. 


Volume  of  any  Solid.     Triple  Integration. 

140.    If  we  suppose  our  sohd  divided  into  parallelopipeds  by 

planes  parallel  to  the  three  coordinate  planes,  the  elementary 

31 


parallelopiped  at  any  point  (x,y,z)  within  the  solid  will  have  for 
its  volume  AccA?/A2,  or,  if  we  regard  x,  y,  and  z  as  independent, 
dxdydz ;  and  the  whole  volume 


17'=  j    I   I  dxdydz, 


[1] 


the  limits  being  so  chosen  as  to  embrace  the  whole  solid. 

The  integrations  are  independent,  and  may  be  performed  in 
any  order  if  the  limits  are  suitably  chosen. 

For  example ;  let  us  find  the  volume  of  the  portion  of  the 
ellipsoid  ^2      ^2      ^2 

cut  off  by  the  coordinate  planes. 


136  INTEGRAL   CALCULUS.  [Art.  140. 

7"=  j    I    I  dzdydx, 

and  our  limits  are,  for  z,  0  and  c\ll ^  —  4^;  for  y,  0  and 

^  \        a^       ¥ 

6-*|l  — — ;    and  for  x,   0  and  a.      For,  starting  at  any  point 

(x,y,z)  and  integrating  on  the  h3'potliesis  that  z  alone  varies,  we 
get  a  column  of  our  elementarj'  parallelopipeds  having  dxdy  as  a 
base  and  passing  through  the  point  {x,y,z).  To  make  this  col- 
umn reach  from  the  plane  XY  to  the  surface,  z  must  increase 
from  the  value  zero  to  the  value  belonging  to  the  point  on  the 
surface  of  the  ellipsoid  which  has  the  coordinates  x  and  y  ;  that 


is,  to  the  value  ca|1 5  —  -^-      Then,  integrating  on  the  h}'- 

\        cr      b- 

pothesis  that  y  alone  varies,  we  shall  sum  these  columns  and 

shall  get  a  slice  of  the  solid  passing  through  {x,y,z)  and  having 

the  thickness,  dx.     To  make  this  slice  reach  completely  across 

the  solid,  we  must  let  y  increase  from  the  value  zero  to  the 

greatest  value  it  can  have  in  the  slice  in  question  ;  that  is,  to  the 

value  which  is  the  ordinate  of  that  point  of  the  section  of  the 

ellipsoid  by  the  plane  XF  which  has  the  abscissae.    The  section 

in  question  has  the  equation 


~2  "■"    12   —        ' 


therefore  the  required  value  of  y  is  h 


>R- 


Last,  in  integrating  on  the  hj^pothesis  that  x  alone  varies,  we 
must  choose  our  limits  so  as  to  include  all  the  sUces  just  de- 
scribed, and  must  increase  x  from  zero  to  a. 


/' 


&=.=caji-5-|; 


between  the  limits     0     and    cH  1  —  — ,  —  ^„ 

^        a-     W 


Chap.  XII.]  VOLUMES.  137 


^U-i-py 


5/"s|»'('l-5)-/-'-!!/ 


nrhn.  / .        (rr\  ^  Ct 


— 1£^  fi  _  ^ 


between  the  limits      0     and     6-\  1 . 


4.1   V        «V  6    ' 


the  vokime  required. 

Examples. 

(1)  Find  the  vokime  obtained  in  tlie  present  article,  perform- 
ing the  integrations  in  the  order  indicated  b}'  the  formula, 


F'=  I    I    I  dxclydz. 


(2)  Find  the  volume  cut  off  from  the  surface 

t^y^^2x 

c^  b 

hj  a  plane  parallel  to  that  of  TZ,  at  a  distance  a  from  it. 

Ans.   •7ra^V(&c), 


(3)  Find  the  volume  enclosed  by  the  surfaces, 

Ans. 


x^  -\-y^  =  cz,     7?  -\-y-  =  ate,     2  =  0.  ^  -'■''' 


32  c 
(4)  Obtain  the  volume  bounded  by  the  surface 


'Z  =  a  —  Va-^  +  y^ 

and  the  planes  x  =  z    and    x  =  0.  Ans. 

^  9 


138 


INTEGRAL   CALCULUS. 


[Art.  141. 


5.   Find  the  volume  of  the  conoid  bounded  by  the  surface 

2      2  nrf^  ft 

2  _^  ^  y  _  g2  ^y^^  j^jjg  planes  x  =  0  and  x  =  a.         Ans.  — -— . 


141.    If  we  use  polar  coordinates  we  can  take  as  our  element 
of  volume 

an  expression  easily  obtained  from  the  element  2 -n-r^  sin  cf>drd(f> 
used  in  Art.  139. 

Then  V=  C  f  Ci^  sin cfidrdcjide, 

where  the  order  of  the  integrations  is  usually  immaterial  if  the 
limits  are  property  chosen. 

Example. 
Find  the  volume  of  a  sphere  by  polar  coordinates. 


Chap.  XIII.] 


CENTRES   OE   GRAVITY. 


139 


CHAPTER   XIII. 


CENTRES    OF    GRAVITY. 


142.  The  moment  of  a  force  about  an  axis  is  the  product  of 
the  magnitude  of  the  force  by  the  perpendicular  distance  of  its 
line  of  direction  from  the  axis,  and  measures  the  tendency  of  the 
force  to  produce  rotation  about  the  axis. 

The  force  exerted  by  gravity  on  any  material  body  is  propor- 
tional to  the  mass  of  the  bod}',  and  may  be  measured  by  the 
mass  of  the  body. 

The  Centre  of  Gravity  of  a  bod}'  is  a  point  so  situated  that  the 
force  of  gravity  produces  no  tendency  in  the  bod}'  to  rotate  about 
an}""  axis  passing  through  this  point. 

The  subject  of  centres  of  gravit}-  belongs  to  Mechanics,  and 
we  shall  accept  the  definitions  and  principles  just  stated  as  data 
for  mathematical  work,  without  investigating  the  mechanical 
gTounds  on  which  they  rest. 


143..  Suppose  the  points  of  a  body  referred  to  a  set  of  three 
rectangular  axes  fixed  in  the  body,  and  let  x,y^z  be  the  coordi- 
nates of  the  centre  of  gi-avit}-.  Place 
the  body  with  the  axes  of  X  and  Z 
horizontal,  and  consider  the  tendenc}'' 
of  the  particles  of  the  body  to  produce 
rotation  about  an  axis  through  (x,y,z) 
parallel  to  OZ,  under  the  influence  of 
gravit}^  Represent  the  mass  of  an 
elementary  paraUelopiped  at  an}-  point 
{x,y,z)  by  dm.  The  force  exerted  by 
gravity  on  dm  is  measured  b}'  dm,  and 

its  line  of  direction  is  vertical.     If  the  mass  of  dm  were  concen- 
trated at  P,  the  moment  of  the  force  exerted  on  dm  about  the 


140  INTEGRAL  CALCULUS.         [Art.  143. 

axis  through  C  would  be  {x  —  x)dm,  and  this  moment  would 
represent  the  tendency  of  dm  to  rotate  about  the  axis  in  ques- 
tion ;  the  tendenc}^  of  the  whole  bod}'  to  rotate  about  this  axis 
would  be  1{x  —  x)dm.  If  now  we  decrease  dm  indefinitely,  the 
error  committed  in  assuming  that  the  mass  of  dm  is  concentrated 
at  P  decreases  indefinitel}',  and  we  shall  have  as  the  true  expres- 
sion for  the  tendenc}'  of  the  whole  bocl}^  to  rotate  about  the  axis 

through  C,   I  (^  —  x)dm  ;  but  this  must  be  zero. 
Hence  i  (x  —  x)dm  =  0, 

I  xdm  —  xi  dm  =  0, 
I  xdm 

^=7^ [1] 

I  dm 

K  we  place  the  body  so  that  the  axes  of  Y  and  X  are  hori- 
zontal, the  same  reasoning  will  give  us 


y  = 


I  ydm 


[2] 


I  dm 

and  in  like  manner  we  can  get 

I  zdm 

^  =  > [3] 

I  dm 

Since  i  dm  is  the  mass  of  the  whole  body,  If  we  represent  it 

by  M  we  shall  have  ^ 

I  xdm 


x  = 


M 

J 

y  = 


I  ydm 


J 


M 

zdm 
M 


Chap.  XIII.]  CENTRES   OF   GRAVITY.  141 


Example. 

Show  that  the  effect  of  gravity  in  making  a  bod}'  tend  to  rotate 
an}-  given  axis  is  precisely'  the  same  as  if  the  mass  of  the  body 
were  concentrated  at  its  centre  of  gravity. 

144.  The  mass  of  any  homogeneous  bod}'  is  the  product  of 
its  volume  b}'  its  density.  If  the  bod}'  is  not  homogeneous,  the 
density  at  any  point  will  be  a  function  of  the  position  of  that 
point.  Let  us  represent  it  by  k.  Then  we  may  regard  dm  as 
equal  to  kcIv  if  civ  is  the  element  of  volume,  and  we  shall  have 

I  XKdv 

^--c —  [1] 

I  Kdv 

and  corresponding  formulas  for  y  and  z. 

If  the  body  considered  is  homogeneous,  k  is  constant,  and  we 
shall  have 

xdv 


i  xdv       I  a 


[2] 


/..      y 


zdv 


[4] 


In  any  particular  problem  we  have  only  to  express  dv  in 
terms  of  the  coordinates. 

Plane  Area. 

145.    If  we  use  rectangular  coordinates,  and  are  dealing  with 
a  plane  area,  where  the  weight  is  uniformly  distributed,  we  have 

dv  =  dA  =  dxdy.  (Art.  125) . 


142  INTEGRAL   CALCULUS. 

Hence,  by  144,  [2]  and  [3], 

j    I  xclxdy 


[Art.  145. 


and 


I    I  dxdy 

j j ydxdy 
,    I   I  dxdy 


If  we  use  polar  coordinates, 

dv  —  dA  =  rd(f>dr, 

I   I  9*^  cos  (f>  d<f)dr 


[1] 


x  = 


y  = 


i    I  rdifidr 

j    I  ?"^  sin  cf}  dcf)dr 
C  Crd4>dr 


[2] 


For  example  ;  let  us  find  the  centre  of  gravity  of  the  area  be- 
tween the  cissoid  and  its  asj^mptote.  From  the  equation  of  the 
cissoid 

r  = » 

a  —  x 

we  see  that  the  curve  is  S3'mmetrical  with  respect  to  the  axis 
of  X,  passes  through  the  origin,  and  has  the  hne  x  =  a  as  an 
asymptote.  From  the  s}Tnmetry  of  the  area  in  question,  y.=  0, 
and  we  need  only  find  x.    ' 


I  I  xdydx       I  xydx 

-  _  Jo     J-y  _  Jo 

I  I  dydx        I  ydx 

Jo  J-y             Jo 


Chap.  XIII.]  CENTRES   OF   GEAVITY.  143 

Jo(a-xy^      _      Jo(«-x).      .    byArt.64[4]. 
Jo{a  —  x)^  Jo(a  — a;)5 


As  an  example  of  the  use  of  the  polar  formulas  [2],  let  us  find 
the  centre  of  gravity  of  the  cardioide 

r=  2a(l— cos^). 

Here,  from  the  fact  that  the  axis  of  X  is  an  axis  of  symmetry, 
we  know  that  y  =  0. 


i '1^  COS  (J3drd<f> 
I       I  rdrd^ 

i  I  r^co&4>d(f>      ——  I  (1  —  eos<f>ycoscf>d(f> 

_  Jo o  Jo 

i  r?-2  d<i>  2a^  C{1-  cos  <t>f  dcji 

X27r 
(cos<^  —  3  cos^^  +  3  cos^c^—  cos*^)dq!)  =  —  -V"^  5 

X27r 
(1  —  2cos^  +  cos-^)cZ0  =  37r. 

Hence  x  =  —  ^a. 

Examples. 

1.  Show  that  formulas  [1]  hold  even  when  we  use  oblique 
coordinates. 

2.  Find  the  centre  of  gravit}^  of  a  segment  of  a  parabola  cut 
off  by  any  chord. 

Ans.   i=|a,     y  =  0.     If  the  axes  are  the  tangent  parallel 
to  the  chord  and  the  diameter  bisecting  the  chord. 


144  INTEGRAL  CALCULUS.         [Art.  146. 

3.  Find  the  centre  of  gravitj"  of  the  area  bounded  b}'  the  semi- 
cubical  parabola  ay'^  =  x^  and  a  double  ordinate.      Ans.  x  =  -f-cc. 

4.  Find  the  centre  of  gravity  of  a  semi-ellipse,  the  bisecting 
line  being  any  diameter. 

Ans.  If  the  bisecting  diameter  is  taken  as  the  axis  of  Y,  and 

4:  (X 

the  conjugate  diameter  as  the  axis  of  X,    x  =  — ,     y  =  0. 

5.  Find  the  centre  of  gravit}^  of  the  curve  y"^  —  b^    ~    . 

X 

Ans.  x==^a. 

6.  Find  the  centre  of  gravity  of  the  cycloid, 

Ans.   X  =  air,    y=^a. 

7.  Find  the  centre  of  gravity  of  the  lemniscate  ?-  =  a-  cos  2  eft. 

.  -        7rV2 

Ans.   x  = a. 

8 

8.  Find  the  centre  of  gravity  of  a  circular  sector. 

A71S.  If  we  take  the  radius  bisecting  the  sector  as  the  axis 

of  X,  and  represent  the  angle  of  the  sector  b}^  a,   ^  =  f 

a 

9.  Find  the  centime  of  gravity  of  the  segment  of  an  ellipse  cut 
off  b}'  a  quadrantal  chord.       Ans.   x  =  ^ -,     y  =  ^ -• 

TT  2  TT  2 

10.  Find  the  centre  of  gravity  of  a  quadrant  of  the  area  of  the 
curve  oji  -f  2/1  =  al.  Ans.  x  =  y  =  fff  -. 

TT 

146.    If  we  are  dealing  with  a  homogeneous  solid  formed  by 
the  revolution  of  a  plane  curve  about  the  axis  of  X,  we  have 

civ  =  2  irydydx.  (Art.  1 38  [  1  ] ) 

Hence,  by  Art.  144  [2], 

I    j  xydxdy 

I   I  ydxdy 


Chap.  XIII.]  CENTE-ES   OF   GEAVITY.  145 

If  we  use  polar  coordinates, 

dv  =  2  TTi^  sin  cl>drdcf>.  (Art.  139  [1]  ) 

i   i  'i^  sin  (^  cos  cf)drd(f> 

Hence  x  =  '^  ^ [2] 

I    1 9-^  sin  (^drdcji 

For  example  ;  let  us  find  the  centre  of  gravit}^  of  a  hemisphere. 
The  equation  of  the  revolving  curve  is  a? -{--if  =  a^. 


fxydydx      ^ 
0    Jo 


If  we  use  polar  coordinates  the  equation  of  the  revolving  curve 
is  r  =  a. 


1      1  r^  sin  rf)  cos  diddtdr      ,    . 
0    Jo  _i«  . 

I      I  1-^  sin  <lid(fidv 


Here         x  = — ^ =  |— g=§a. 

-g-a 


Examples. 

1 .  Find  the  centre  of  gravity  of  the  solid  formed  by  the  revolu- 
tion of  the  sector  of  a  circle  about  one  of  its  extreme  radii. 

Ans.  x  =  ^a  cos"|^|S,  where  /5  is  the  angle  of  the  sector. 

2.  Find  the  centre  of  gravity  of  the  segment  of  a  paraboloid 
of  revolution  cut  off  b}'  a  plane  perpendicular  to  the  axis. 

Ans.  S  =  f  a,  where  x  =  a  is  the  plane. 

3.  Find  the  centre  of  gravity  of  the  solid  formed  by  scooping 
out  a  cone  from  a  given  paraboloid  of  revolution,  the  bases  of 
the  two  volumes  being  coincident  as  well  as  their  vertices. 

Ans.  The  centre  of  gravity  bisects  the  axis. 


146  INTEGRAL  CALCULUS.         [Art.  147. 

4.  A  cardioicle  is  made  to  revolve  about  its  axis ;   find  the 
centre  of  gravity'  of  the  sohd  generated.  Ans.  5;  =  —  |a, 

o.   Obtain  formulas  for  the  centre  of  gravity'  of  an}"  homo- 
geneous solid. 

6.   Find  the  centre  of  gravit}'  of  the  solid  bounded  b}'  the 
surface  z-  —  xy  and  the  five  planes  cc=0,  y=0,  2=0,  x=a,  y=h. 

Ans.   a;  =  |a,  ^=|&,   z^^a^hl. 

147.    If  we  are  dealing  with  the  arc  of  a  plane  curve,  the 
formulas  of  Art.  144  reduce  to 

I  xds 
3!  =  <^,  [1] 


Examples. 

1.  Find  the  centre  of  gravity-  of  an  arc  of  a  circle,  taking  the 
diameter  bisecting  the  arc  as  the  axis  of  X  and  the  centre  as  the 

origm.  Ans.  x  =  — ,  where  c  is  the  chord  of  the  arc. 

s 

2.  Find  the  centre  of  gravity-  of  the  arc  of  the  curve  fljl-|-?/l=as 
between  two  successive  cusps.  Ans.  x  —  y  =  ^a. 

3.  Find  the  centre  of  gravity  of  the  arc  of  a  semi-cjxloid. 

Ans.  .T  =  (7r  — f)a,    ?/  =  — fa. 

4.  Find  the  centre  of  gravity  of  the  arc  of  a  catenar}'  cut  off 
by  any  double  ordinate. 

Ans.  x  —  0,    y  —  i     where  2  s  is  the  length  of  the  arc. 

5.  Obtain  formulas  for  the  centre  of  gravity  of  a  surface  of 
revolution,  the  weight  being  uniformly  distributed  over  the 
surface. 


Chap.  XIII.]  CENTRES   OF   GRAVITY.  147 

6.  Find  the  centre  of  gi-avitj  of  an}-  zone  of  a  sphere. 

Ans.  The  centre  of  gTavity  bisects  the  hne  joining  the  centres 
of  the  bases  of  tlie  zone. 

7.  A  cardioicle  revolves  about  its  axis  ;  find  the  centre  of 
gravity  of  the  surface  generated.  Ans.  x  =  —  -VV-o. 

8.  Find  tlie  centre  of  gravity  of  the  surface  of  a  heinisphere 
when  the  density  at  each  point  of  the  surface  varies  as  its  per- 
pendicular distance  from  the  base  of  the  hemisphere. 

Ans.  X  =  fa. 

9.  Find  the  centre  of  gi-avity  of  a  quadrant  of  a  circle,  the 
density  at  an}'  point  of  which  varies  as  the  nth  power  of  its 
distance  from  the  centre.  ^,^5^  ^^.  _  ^  __  ^?  -h'2  2  a 

n-{-3    TT 

10.  Find  the  centre  of  gravity  of  a  hemisphere,  the  density 
of  which  varies  as  the  distance  from  the  centre  of  the  sphere. 

Ans.  x  =  fa. 

Properties  of  Giddin. 

148.  I.  If  a  plane  area  revolve  about  an  axis  external  to 
itself  through  any  assigned  angle,  the  volume  of  the  solid  gene- 
rated will  be  equal  to  a  prism  whose  base  is  the  revolving  area 
and  whose  altitude  is  the  length  of  the  path  described  by  the 
centre  of  gravity-  of  the  area. 

11.  If  the  arc  of  a  plane  curve  revolve  about  an  external  axis 
in  its  own  plane  through  any  assigned  angle,  the  area  of  the 
surface  generated  will  be  equal  to  that  of  a  rectangle,  one  side 
of  which  IS  the  length  of  the  revohang  curve,  and  the  other  the 
length  of  the  path  described  b}'  its  centre  of  gravit}-. 

First ;  let  the  area  in  question  revolve  about  the  axis  of  X 
through  an  angle  ©.  The  ordinate  of  the  centre  of  gravity  of 
the  area  m  question  is 

I    I  ydxdy 

y--^ .  by  Art.  145  [1]. 

I   I  dxdy 


148  iNTEGEAIi   CALCULUS.  [Art.  148. 

The  length  of  the  path  described  by  the  centre  of  gravity 

0  j    j  ydxcly 

I    I  dxdy 
The  vohime  generated  is 

F=    ®CCydxdy,  by  Art.  138. 

Hence  V=y®iidxdy. 

But  I    I  dxdy  is  the  revolving  area,  and  the  first  theorem  is 

established. 

We  leave  the  proof  of  the  second  theorem  to  the  student. 

Examples. 

1.  Find  the  surface  and  volume  of  a  sphere,  regarding  it  as 
generated  by  the  revolution  of  a  semicircle. 

2.  Find  the  surface  and  volume  of  the  solid  generated  by  the 
revolution  of  a  cycloid  about  its  base. 

3.  Find  the  volume  and  the  surface  of  the  ring  generated  by 
the  revolution  of  a  circle  about  an  external  axis. 

Ans.    F=27r^a^6,     S  =  A7rhib,    where  b  is  the  distance  of 
the  centre  of  the  circle  from  the  axis. 

4.  Find  the  volume  of  the  ring  generated  by  the  revolution  of 
an  ellipse  about  an  external  axis. 

Ajis.   F=  7r^a5c,  where  c  is  the  distance  of  the  centre  of  the 
ellipse  from  the  axis. 


Chap.  XIV.]      MEAN   VALUE   AND   PEOB ABILITY.  149 


CHAPTER    XIV. 
MEAN  VALUE  AND   PEOBABILITY. 

149.  The  application  of  the  Integral  Calculus  to  questions 
in  Mean  Value  and  Probability  is  a  matter  of  decided  interest ; 
but  lack  of  space  will  prevent  our  doing  more  than  solving  a 
few  problems  in  illustration  of  some  of  the  simplest  of  the  meth- 
ods and  devices  ordinaril}^  employed.  A  full  anfl  admirable 
treatment  of  the  subject  is  given  in  "Williamson's  Integral 
Calculus"  (London:  Longmans,  Green,  &  Co.)  ;  and  numer- 
ous interesting  problems  are  published  with  then*  'solutions  in 
"The  Mathematical  Visitor,"  a  magazine  edited  by  Artemas 
Martin,  Erie,  Pa. 

150.  The  mean  of  n  quantities  is  their  suon  divided  by  their 
member.  If  we  are  finding  the  mecm  value  of  a  continuously- 
varj'ing  quantity,  we  have  to  consider  an  infinite  number  of 
values,  and,  of  course,  an  infinite  sum  as  well ;  a  Httle  ingenuity 
will  enable  us  to  throw  the  ratio  of  the  sum  to  the  number  into 
a  form  to  which  we  can  apply  the  Integral  Calculus. 

(a)  Let  us  find  the  mean  distance  of  all  the  points  on  the 
circumference  of  a  cu'cle  from  a  given  point  on  the  cu-cumfer- 
ence. 

If  we  take  the  given  point  as  origin,  the  distances  whose 
mean  is  required  are  the  radii  vectores  of  points  uniformly  dis- 
tributed along  the  circumference  of  the  circle. 

Let  the  distance  between  any  two  adjacent  points  be  ds  ;  then 

w,  the  number  of  points  considered,  is  equal  to  ,  if  a  is  the 

ds 


150  INTEGRAL   CALCULUS.  [Art.  150. 

radius  of  the  circle  ;  and  if  r  is  the  radius  vector  of  am'  point  of 
the  cu'cumference,  and  7lif  the  mean  vahie  required, 

M=  —  =  ^^ 
2  Tva      2  ira 

ds 

for  any  finite  value  of  ds.     In  the  actual  case  under  considera- 
tion, 

I  rds 

27ra 
The  polar  equation  of  the  circle  is 

r  =  2a  cos ^ ; 
ds=  2ad(ji, 


M  = I    4  a^  cos  (jid(ji  =  — , 


the  required  mean  value. 

(b)  Let  us  find  the  mean  distance  of  points  on  the  surface 
of  a  circle  from  a  fixed  point  on  the  circumference. 
Using  the  same  notation  as  before,  we  shall  have 


rdcfidr 


r  and  <^,  the  polar  coordinates  of  any  point  of  the  surface,  being 
independent  variables. 

M= T-= 5-^? 


rdrdcji 

ivr        1    n  C'^rA      32  a 
Jf=—         lr-dnlcf,  =  -—, 

Tra'^.l^n  y  TT 

2 

the  required  mean  value. 


Chap.  XIV.]  MEAInT  VALUE   AND   PKOBABILITY.  151 

(c)  As  an  example  of  a  device  often  emploj-ed,  we  shall  now 
solve  tlie  problem,  To  find  the  mean  distance  between  two  points 
within  a  given  circle. 

If  M  be  the  required  mean,  the  sum  of  the  whole  number  of 
cases  can  be  represented  b}'  {Trf-)-M,  r  being  the  radius  of  the 
circle  ;  since  for  each  position  of  the  first  point  the  number  of 
positions  of  the  second  point  is  proportional  to  the  area  of  the 
circle,  and  may  be  measured  by  that  area ;  and  as  the  number 
of  possible  positions  of  the  first  point  may  also  be  measured 
by  the  area  of  the  cu'cle,  the  whole  number  of  cases  to  be  con- 
sidered is  represented  by  the  square  of  the  area ;  and  the  sum 
of  all  the  distances  to  be  considered  must  be  the  product  of  the 
mean  distance  b}'  the  number. 

Let  us  see  what  change  will  be  produced  in  this  sum  b}^  in- 
creasing r  by  the  infinitesimal  dr  ;  that  is,  let  us  find  d(7rh*M). 

If  the  first  point  is  anj'where  on  the  annulus  2  Trr.dr,  which  we 
have  just  added,  its  mean  distance  from  the  other  points  of  the 

cu'cle  IS ,  by  (o). 

Therefore,  the  sum  of  the  new  distances  to  be  considered, 

39  J. 
if  the  first  point  is  on  the  annulus,  is  -^^ .  ir')^ .  2 -n-rdr ;  but  the 

9  TT 

second  point  may  be  on  the  annulus,  instead  of  the  first ;  so  that 
to  get  the  sum  of  all  the  new  cases  brought  in  by  increasing 
r  by  dr,  we  must  double  the  value  just  obtained. 

Hence  d(Trr'M)  =  ^s.^r\lr, 


45  TT 


151.  In  solving  questions  in  Probability,  we  shall  assume 
that  the  student  is  familiar  with  the  elements  bf  the  theory  as 
given  in  "  Todhunter's  Algebra." 

(a)  A  man  starts  from  the  bank  of  a  straight  river,  and 
wallis  till  noon  in  a  random  du*ection  ;  he  then  turns  and  walks 


152  INTEGRAL  CALCULUS.  [Art.  151. 

in  another  random  direction  ;  what  is  the  probability  that  he  will 
reach  the  river  by  night  ? 

Let  6  be  the  angle  his  first  course  makes  with  the  river.  If 
the  angle  through  which  he  turns  at  noon  is  less  than  vr  —  2  ^, 
he  will  reach  the  river  by  night.     For  any  given  value  of  6, 

then,  the  required  probability  is  .     The  probability  that 

9  shall  lie  between  any  given  value  6q  and  ^o  +  c^^  is  — . 

The  chance  that  his  first  course  shall  make  an  angle  with  the 
river  between  Bq  and  9q  +  dO,  and  that  he  shall  get  back,  is 

7r-2g    cie  _  {7r-2e)cl6 


2 


TT  -tTT 


As  6  is  equall}^  likety  to  have  an}^  value  between  0  and—,  the 
required  probability, 


2 


.\7r-26)d6_, 
P—    I  I  —  4- 

^0 


(5)  A  floor  is  ruled  with  equidistant  straight  lines ;  a  rod, 
shorter  than  the  distance  between  the  lines,  is  thrown  at  ran- 
dom on  the  floor  ;  to  find  the  chance  of  its  falling  on  one  of  the 
lines. 

Let  X  be  the  distance  of  the  centre  of  the  rod  from  the  nearest 
line  ;  $  the  inclination  of  the  rod  to  a  perpendicular-to  the  paral- 
lels passing  through  the  centre  of  the  rod ;  2  a  the  common  dis- 
tance of  the  parallels  ;  2  c  the  length  of  the  rod. 

Li  order  that  ,the  rod  may  cross  a  line,  we  must  have 
c cos 6>  x;  the  chance  of  this  for  any  given  value  a-Q  of  x  is 

—  cos  ^  — . 

The  probability  that  x  will  have  the  value  x^  is  — .  The 
probability  required  is 

iX  .        2c 


*  C        X 

Jq  c 


■n-a 


This  problem  ma^j  be  solved  by  another  method  which  pos- 
sesses considerable  interest. 


Chap.  XIV.]  MEAN   VALUE   AND   PEOBABILITY.  153 

Since  all  values  of  x  from  0  to  a,  and  all  values  of  6  from  —  | 
to  -  are  equally  probable,  the  whole  number  of  cases  that  can 
arise  ma}'  be  represented  b}- 

P"  Cdxde  = 


TTCI. 


The  number  of  favorable  cases  will  be  represented  b}' 

\ 

.c  COS  9 


(        \  dxcW  =  2i 


2c 
Hence  p  =  — 

TTtt 

(c)  To  find  the  probability  that  the  distance  of  two  stars, 
taken  at  random  in  the  northern  hemisphere,  shall  exceed  90°. 

Let  a  be  the  latitude  of  the  first  star.  With  the  star  as  a 
pole,  describe  an  arc  of  a  great  circle,  dividing  the  hemisphere 
into  two  lunes  ;  the  probability  that  the  distance  of  the  sec- 
ond star  from  the  first  will  exceed  90°  is  the  ratio  of  the  lune 
not  containing  the  first  star  to  the  hemisphere,  and  is  equal 
to  vr^LZL^.     The  probability  that  the  latitude  of  the  first  star 

will  be  between  a  and  a  +  da  is  the  ratio  of  the  area  of  the 
zone,  whose  bounding  circles  have  the  latitudes  a  and  a  -\-  da. 
respectivel}',  to  the  area  of  the  hemisphere,  and  is 

2  Tra-  cos  a  da  ■, 
=  cos  a  aa. 

2  TTcr 

a)  ,  1 


Hence  p=  I  "il^I — ^  cos  a  da 

•^0  TT 


(d)  A  random  straight  line  meets  a  closed  -convex  curve  ; 
what  is  the  probability  that  it  will  meet  a  second  closed  convex 
curve  within  the  first  ? 

If  an  infinite  number  of  random  lines  be  drawn  in  a  plane,  all 
du-ections  are  equally  probable  ;    and  lines  having  any  given 


154:  .  INTEGEAL  CALCULUS.         [Art.  151. 

direction  will  be  disposed  with  equal  frequenc}'  all  over  the 
plane.  If  we  determine  a  line  b}'  its  distance  p  from  the  origin, 
and  b}'  the  angle  a  which  p  makes  with  the  axis  of  X,  we  can  get 
all  the  lines  to  be  considered  b}'  making  p  and  a  vary  between 
suitable  limits  b}'  equal  infinitesimal  increments. 

In  our  problem,  the  whole  number  of  lines  meeting  the  exter- 
nal curve  can  be  represented  by   |    |  clpda.     If  the  origin  is 

within  the  curve,  the  limits  for  p  must  be  zero,  and  the  perpen- 
dicular distance  from  the  origin  to  a  tangent  to  the  curve  ;  and 
for  a  must  be  zero  and  2  7r.  If  we  call  this  number  N,  we 
shall  have 

N  = 


pda^ 

0 


p  being  now  the  perpendicular  from  the  origin  to  the  tangent. 

If  we  regard  the  distance  from  a  given  point  of  any  closed 
convex  curve  along  the  curve  to  the  point  of  contact  of  a  tan- 
gent, and  then  along  the  tangent  to  the  foot  of  the  perpendicu- 
lar let  fall  upon  it  from  the  origin,  as  a  function  of  the  a  used 
above,  its  differential  is  easil}'  seen  to  be  pda.  If  we  sum  these 
differentials  from  a  =  0  to  a  =  2  tt,  we  shall  get  the  perimeter 
of  the  given  curve. 


Hence  -ZV 


=  i  pda  =  L, 


where  L  is  the  perimeter  of  the  curve  in  question.  B}'  the  same 
reasoning,  we  can  see  that  %,  the  number  of  the  random  lines 
which  meet  the  ifiner  curve,  is  equal  to  Z,  its  perimeter.  For  p, 
the  required  probability,  we  shall  have 

I 
^  =  L 

Examples. 

(1)  A  number  n  is  divided  at  random  into  two  parts  ;  find  the 
mean  value  of  their  product.  ,^^^^    ^ 

6* 


Chap.  XIV.]  MEAN   VALUE   AND    PROBABILITY.  155 

(2)  Find  the  mean  value  of  the  ordinates  of  a  semicircle,  sup- 
posing the  series  of  ordinates  taken  equidistant.  Ans.  -a. 

4 

(3)  Find  the  mean  A'alue  of  the  ordinates  of  a  semicircle,  sup- 
posing the  ordinates  drawn  through  equidistant  points  on  the 
circumference.  a        2a 

TV 

(4)  Find  the  mean  values  of  the  roots  of  the  quadratic 
X-  —  ax  -f-  &  =  0,  the  roots  being  known  to  be  real,  but  b  being 
unknown  but  positive.  a    ,    5a       -,  a 

6    "        6" 

(5)  Prove  that  the  mean  of  the  radii  vectores  of  an  ellipse,  the 
focus  being  the  origin,  is  equal  to  half  the  minor  axis  wheh  they 
are  di'awn  at  equal  angular  intervals,  and  is  equal  to  half  the 
major  axis  when  the}'  are  drawn  so  that  the  abscissas  of  their 
extremities  increase  uniformly. 

(6)  Suppose  a  straight  line  divided  at  random  into  three 
parts  ;  find  the  mean  value  of  their  product.  -         a^ 

60 " 

(7)  Find  the  mean  square  of  the  distance  of  a  point  within  a 
given  square  (side  =  2  a)  from  the  centre  of  the  square. 

A71S.  f  a^. 

(8)  A  slab  is  sawed  at  random  from  a  round  log,  find  its 

mean  tliickness.  .        4  a 

Ans.  — . 

o 
OTT 

(9)  A  chord  is  drawn  joining  two  points  taken  at  random  on 

a  circle  ;  find  the  mean  area  of  the  less  of  the  two  segments  into 

which  it  divides  the  cii'cle.  ^         7ra^      a^' 

Ans. 

4  TT 

(10)  Find  the  mean  latitude  of  all  places  north  of  the  equator, 

Ans.    32°. 7. 

(11)  Two  points  are  taken  at  random  in  a  triangle  ;  find  the 
mean  area  of  the  triangular  portion  which  the  line  joining  them 
cuts  off  from  the  whole  triangle.  Ans.    |  of  the  whole. 

(12)  Find  the  mean  distance  of  points  within  a  sphere  from  a 
given  point  of  the  surface.  Ans.    |a. 


156  INTEGRAL  CALCULUS.         [Art.  151. 

(13)  Find  the  mean  distance  of  two  points  taken  at  random 
within  a  sphere.  Ans.    |fa. 

(14)  Two  points  are  taken  at  random  in  a  given  line  a  ;  find 
the  chance  that  their  distance  shall  exceed  a  given  value  c. 

^a  —  c\- 


Ans. 


a 

(15)  Find  the  chance  that  the  distance  of  two  points  within  a 
square  shall  not  exceed  a  side  of  the  square.  Ans.   tt  —  -^^-. 

(16)  A  line  crosses  a  circle  at  random  ;  find  the  chances  that 

a  point,  taken  at  random  within  the  circle,  shall  be  distant  from 

the  line  by  less  than  the  radius  of  the  circle.  .         .       2 

•^  Ans.    1  —  77-- 

OTT 

(17)  A  random  straight  line  crosses  a  circle  ;  find  the  chance 
that  two  points,  taken  at  random  in  the  circle,  shall  lie  on  oppo- 
site sides  of  the  line.  .         128 

Ans. 

45  TT^ 

(18)  A  random  straight  line  is  drawn  across  a  square  ;   find 

the  chance  that  it  intersects  two  opposite  sides.  loo- 2 

^  A71S.  i ^. 

(19)  Two  arrows  are  sticking  in  a  circular  target;  find  the 
chance  that  then-  distance  apart  is  greater  than  the  radius. 


Ans. 


3V3 


47r 

(20)  From  a  point  in  the  circumference  of  a  circular  field  a 
projectile  is  thrown  at  random  with  a  given  velocity  which  is 
such  that  the  diameter  of  the  field  is  equal  to  the  greatest  range 
of  the  projectile ;  ^nd  the  chance  of  its  falling  within  the  field. 

Ans.   i--(V2-l). 

TT 

(21)  On  a  table  a  series  of  equidistant  parallel  lines  is  drawn, 
and  a  cube  is  thrown  at  random  on  the  table.  Supposing  that 
the  diagonal  of  the  cube  is  less  than  the  distance  between  con- 
secutive straight  lines,  find  the  chance  that  the  cube  will  rest 
without  covering  any  part  of  the  lines. 

Ans.  1 ,  where  a  is  the  edge  of  the  cube  and  c  the  dis- 

ttC 

tance  between  consecutive  lines. 


Chap.  XIV.]  MtlAN  VALUE  AND   PROBABILITY.  157 

(22)  A  plane  area  is  ruled  with  equidistant  parallel  straight 
lines,  the  distance  between  consecutive  lines  being  c.  A  closed 
curve,  having  no  singular  points,  whose  greatest  diameter  is  less 
than  c,  is  thi'own  down  on  the  area.  Find  the  chance  that  the 
curve  falls  on  one  of  the  lines. 

Alls.  — ,  where  I  is  the  perimeter  of  the  curve. 

TTC 

(23)  During  a  heavy  rain-storm,  a  circular  pond  is  formed  in 
a  circular  field.  K  a  man  undertakes  to  cross  the  field  in  the 
dark,  what  is  the  chance  that  he  will  walk  into  the  pond? 


158  INTEGEAL   CALCULIJS.  [Art.  152. 


CHAPTER    XV. 

KEY  TO   THE   SOLUTION   OF  DIFFEEENTIAL   EQUATIONS. 

152.  In  this  chapter  an  analytical  ke}-  leads  to  a  set  of  con- 
cise, practical  rules,  embod^dng  most  of  the  ordinary  metliods 
emplo3'ed  in  solving  differential  equations  ;  and  the  attempt  has 
been  made  to  render  these  rules  so  explicit  that  they  may  be 
understood  and  applied  by  any  one  who  has  mastered  the  Inte- 
gral Calculus  proper. 

The  key  is  based  upon  ' '  Boole's  Differential  Equations " 
(London  :  Macmillan  &  Co.),  to  which  the  student  who  wishes 
to  become  familiar  with  the  theoretical  considerations  upon 
which  the  working  rules  are  based  is  referred. 

153.  A  differential  equation  is  an  expressed  relation  involv- 
ing derivatives  with  or  without  the  primitive  variables  from 
which  they  are  derived. 

For  example : 

{l  +  x)y  +  {\-y)x^^  =  Q,  (1) 

«'|-«2/  =  ^'+l.  (2) 

D^z-aWy^z  =  0,  (4) 

are  differential  equations. 

The  order  of  a  differential  equation  is  the  same  as  that  of  the 
derivative  of  highest  order  which  appears  in  the  equation. 

Equations  (1)  and  (2)  are  of  the  first  order;  (3)  and  (4)  of 
the  second  order. 

The  degree  of  a  differential  equation  is  the  same  as  the  power 


Chap.  XV.]  DIFFEEEISTTIAL  EQUATIONS.      KEY.  159 

to  which  the  derivative  of  highest  order  in  the  equation  is  raised, 
that  derivative  being  supposed  to  enter  into  the  equation  in  a 
rational  form. 

Equations  (1),  (2),  (3),  and  (4)  are  all  of  the  first  degree. 

A  differential  equation  is  linear  when  it  would  be  of  the  first 
degree  if  the  dependent  variable  and  all  its  derivatives  wei'e 
regarded  as  unknown  quantities. 

Equations  (2),  (3),  and  (4)  are  linear. 

The  equation  not  containing  differentials  or  derivatives,  and 
expressing  the  most  general  relation  between  the  primitive  vari- 
ables consistent  with  the  given  differential  equation,  is  called 
its  general  solution  or  complete  primitive.  A  general  solution 
will  always  contain  arbitrary  constants  or  arbitrary  func- 
tions. 

A  singular  solution  of  a  differential  equation  is  a  relation  be- 
tween the  primitive  variables  which  satisfies  the  differential 
equation  b}^  means  of  the  values  which  it  gives  to  the  deriva- 
tives, but  which  cannot  be  obtained  from  the  complete  primitive 
by  giving  particular  values  to  the  arbitrary  constants. 

154.  We  shall  illustrate  the  use  of  the  ke}'  by  solving  equa- 
tions (1),  (2),  (3),  and  (4)  of  Art.  157  b}'  its  aid. 

(1)   (l+x)y-\-(l-y)x^=0,  or  (l-hx)yclx-\-(l-y)xdy=0. 

Beginning  at  the  beginning  of  the  key,  we  see  that  we  have  a 
single  equation,  and  hence  look  under  I.,  p.  163  ;  it  involves 
ordinary  derivatives:  we  are  then  directed  to  II.,  p.  163;  it 
contains  two  variables  :  we  go  to  III.,  p.  163 ;  it  is  of  the  first 
order,  IV.,  p.  163,  and  of  the  first  degree,  V.,  p.  163. 

It  is  reducible  to  the  form 


— ! —  dx  -\ dy  =  0, 

X  y 

which  comes  under  Xdx  -\-  Ydy  =  0. 


160  INTEGRAL  CALCULUS.         [Art.  154. 

Hence  we  turn  to  (1),  p.  166,  and  there  find  the  specific  direc- 
tions for  its  solution.     Integrating  each  term  separately,  we  get 

'\ogx  +  x  +  \ogy-y  =  c,     or     log{xy) -^  x  —  y  =  c, 

the  required  primitive  equation, 

(2)  x-£-ay  =  x-hl. 

Beginning  again  at  the  beginning  of  the  key,  we  are  directed 
through  I.,  11.,  III.,  IV.,  to  V.,  p.  163.  Looking  under  V., 
we  see  that  it  will  come  under  either  the  third  or  the  fourth 
head.  Let  us  try  the  fourth;  we  are  referred  to  (4),  p.  167, 
for  specific  directions. 

Obeying  instructions,  the  work  is  as  follows  : 


dx 

-ay  = 

=  0, 

xdy  — 

aydx  = 

=  0, 

dy_ 

adx 

:0, 

y  ' 

X 

lO; 

gy-a 

;logfl;  = 

:C,. 

^"4- 

=  c; 

=  0, 

y 

=  Cx% 

dy 
dx 

=  aCxf 

X-l  _,_  j^< 

^dG 
dx' 

(1) 


Substitute  in  the  given  equation 


rlC 

aCx"  +  a;"+i—  —  aCx"  =  x  +1, 
dx 

dx 
dC~^^dx  =  0, 

fy  J 1 I    _1_  __  Ql 

(a-l)cc«-^     ax" 


Chap.  XV.]  DIFFEEEISTTIAL  EQUATIONS.      KEY.  161 

Substitute  this  value  for  C  iu  (1),  and  we  get 


^a      a  —  1 
the  requu-ed  primitive. 

(3)  |^+H;^*=o. 

dxr       clx 

Beginning  at  the  beginning  of  the  key,  we  are  directed 
tlirough  I.,  II.,  VII.,  to  (20),  p.  171,  for  our  specific  instruc- 
tions. 

Obej'ing  these,  our  work  is  as  follows  : 

y  =  Ce"^, 
dy  =  mCe'"'°  clx, 

Substitute  in  the  equation,  and 

or  m^  +  2  m  =  0  ; 

m  =  0     or     —2. 

y  =  C+C'e-'=' 
is  the  solution  required. 

(i)  JD,'z-a-Dfz  =  0. 

Beginning  at  the  beginning  of  the  kej',  we  are  directed 
through  I.  and  IX.  to  (43),  p.  179,  for  our  specific  instruc- 
tions. 

Obeying  these,  our  work  is  as  follows  : 

dif  —  a-dx^  =  0, 

dy  —adx   =0,  (1) 

dy  -f-  adx   =  0,  (2) 

djxly  —  ahlqdx  =  0.  (3) 

Combining  (1)  and  (3),  we  get 

dpdy  —  adqdy  =  0, 
or  dp  —  adq  =  0.  (4) 


162  INTEGRAL  CALCULUS.         [Art.  154. 

(1)  gives  y  —  ax  =  a. 
(4)  gives                          p  —  aq  =  ^. 

(2)  and  (3)  give  us,  in  the  same  way, 

y  +  ax  —  tti, 

and  our  two  first  integrals  are 

p-aq=f^{y-ax),  (5) 

p  +  aq=f^{y  +  ax),  (6) 

/i  and/2  denoting  arbitrar}-  functions. 
Determining  p  and  g,  from  (5)  and  (6) , 

i>  =  i    [/2(2/  +  «aj)  +/i(2/  -  "a') ] ' 
^  = --[/2(2/ +  aa;) -/i(?/ -  ao;)]  ; 

dz=^^\_f2{y+ax)  +/i(y-ax)]  dcc+— [/2(2/+acc)  +/i(2/-aa;)]d?/ 

f\{y  +  ax)  ((^?/  +  adx)  —f\{y  —  ax)  (dy  —  adx) 
"^  2a 

Hence,  z  =  F(y  +  aa;)  +  i^i  (2/  —  ax) , 

where  F  and  F^  denote  arbitrary  functions  obtained  b}'  integrat- 
ing /i  and  /2,  which  are  arbitrary. 


——* 

Page 

Single  equation .  I.  163 

S^'stem  of  simultaneous  equations     ....     VIII.  166 

I.   luA^olving  ordiuaiy  derivatives II.  163 

Involving  partial  derivatives IX.  166 

II.   Containing  two  variables III.   163 

Containing  three  variables  and  of  first  degree. 

General  form,  Pdx  +  Qdy  +  Bdz  =  0  .     .     .  (34)  1 75 

Containing  more  than  three  variables  and  of 
the  first  degree.     General  form,  Pdxi  -^Qdx2 

+  Bdxs-h =0 (35)176 

III.  Of  first  order IV.  163 

Not  of  first  order VII.  165 

IV.  Of  first  degree.    General  form,  3Idx+]Sfdy=0,  V.  163 
Not  of  first  degree VI.  164 

V.       Of  first  degi-ee.     General  form,  Mdx-{-Ndy  ^   ll) 

=  0. 
Of  or  reducible  to  the  form  Xdx  +  Ydy  =  0, 

where  X  is  a  function  of  x  alone,  and  Y  is  / 

a  function  of  2/ alone  * (1)  166 

M  and  ^  homogeneous  functions  of  x  and  y  of 

the  same  degree (2)  166 

Of  the  form  {ax-\-by-\-c)dx -\-(a'x-^b'y-{-c')dy 

=  0 (3)  167 

Linear.    General  form  ~+Xiy=X2,  where X^ 

and  Xg  are  functions  of  a;  alone  *  ....         (4)  167 

*  Of  course,  X,  X^,  X,,  and  Fmay  be  constants. 

163 


164  INTEGEAL    CALCULUS. 

dv  ^*^* 

Of  the  form  -^  +Xi?/=Xo?/",  where  Xj  and  X2 
dx 

are  functions  of  x  alone* (5)167 

Mdx+Ndy  an  exact  differential.    Test,  DyM 

=  D^N (6)  167 

3Ix  +  Ny  =  0 (7)168 

Mx-Ny  =  0 (8)  168 

Of  i\iQ  form.  F^{xy)ydx  +  F2{xy)xdy  =  0    .     .  (9)168 

DyM-D^N  ^  fjjjjction  of  x  alone    ....       (10)  168 

D.N- DyM  ^  f^j^g^ioj^  of  y  alone    .     .     .     .       (11)  168 

DyM-D,N      f^^^tioj^  of  rr^y\ (12)  168 

Ny-Mx  '                         ^  -^^       .  ^     ^ 

xHD^N-D,M)  +  nNx^   a    function    of  h 
Mx  +  Ny                                           X 
n  being  any  number (13)  168 


VI.       Not  of  first  degree. 

Can  be  solved  as  an  algebraic  equation  in  j3, 

where  p  stands  for  -^ (14)  169 

dx 

Involves   onl}^   one   of  the   variables    and  p, 

(Li! 

where  p  stands  for  -±    . (15)  169 

dx 

Of  the  form  xf^p  +  yf2p  =fsP,  where  2^  stands 

for  ^    '.     .     .     .     . (16)  169 

dx 

Homogeneous  relatively  to  x  and  y   .     .     .     .       (17)  170 

Of  the  form  i^((^,i/f)  =  0,  where  ^  and  ij/  are 

dv 
functions  of  x,  y,  and  — ,  such  that  <^  =  a 
dx 

il/  =  b  will  lead,   on  differentiation,   to  the 
same    differential   equation    of  the    second 

order (18)  170 

A  singular  solution  wiU  answer    .     .     .     ..       (19)1 70 

*  See  note,  p.  163. 


KEY.  165 

VII.       Not  of  first  order.  ^^=* 

Linear,  with    constant    coefficients ;    second 

member  zero* (20)  171 

Linear,  witli    constant    coefficients ;    second 

member  not  zero* (21)  171 

Of  the  form  (a+6x-)"^+^(a+&a;)"-^?^ 

+ -\-  Ly  =  X,  where  X  is   a  function 

of  X  alonet (22)  172 

Either  of  the  primitive  variables  wanting      .       (23)  172 

cZ"?/ 
Of  the  form  — ^  =  X,  X  being  a  function  of 

X  alonet (24)  172 

Of  the  form  —4  =Y,   Y  being  a  function  of 

?/ alone  t (25)  172 

Of  the  form  ^=/^^       .......  (26)172 

In  In— 2 

Of  the  form  ^=/^^,    • (27)173 

dx"     -^  dx"-  ^     ' 

Homogeneous  on  the  supposition  that  x  and 
xj  are  of  the  degree  1,  -^  of  the  degree  0, 
^  of  the  degree  -1, (28)173 

Homogeneous  on  the  supposition  that  x  is  of 
the  degree  1 ,  ?/  of  the  degree  w,  -^  of  the 

degree  w  —  1,  ^  of  the  degree  n  —  2,  ,       (29)  173 

dx-  ,       ,9 

Homogeneous  relativel}'  to  ?/,  — ,  -^,  .       (30)  173 

Containing  the  first  power  onl}'  of  the  deriva- 
tive of  the  highest  order (31)  173 


Of  the  form  ^,  +X$^  +  T 
dx-         dx 


0,  where 


~dy_ 
dx 

X  is  a  function  of  x  alone  and  Y  a  func- 
tion of  y  alonet (32)  174 

Singular  integral  will  answer      (33)  174 

*  The  first  member  is  supposed  to  contain  only  those  terms  involving  the  depen- 
dent variable  or  its  derivatives. 
t  See  note,  p.  163. 


166  INTEGEAL   CALCULUS. 

Page 

VIII.  Simultaneous  equations  of  the  first  order  .     .       (36)  176 
Not  of  the  first  order (37)  177 

IX,  All  the  partial  derivatives  taken  with  respect 

to  one  of  the  independent  variables       .     .        (38)  177 

Of  the  first  order  and  Linear X.  166 

Of  the  first  order  and  not  Linear  ....  XL  166 
Of  the  second  order  and  containing  the  deriv- 
atives of  the  second  order  onl}^  in  the  first 
degree.  General  form  RD^z  -\-  SD^DyZ  + 
TDy^z  =  V,  where  B,  S,  T,  and  V  may  be 
functions  of  x,  y,  z,  D^z,  and  D^z  .     .     .       (43)  179 

X.  Containing  three  variables (39)  178 

Containing  more  than  three  variables  .     .     .       (40)  178 

XI.  Containing  three  variables (41)  178 

Containing  more  than  three  variables  .     .     .       (42)  179 


(1)  Of  or  reducible  to  the  form  Xdx  +  Ycly  =  0,  where  X  is 
a  function  of  x  alone  and  Y"  is  a  function  of  y  alone. 

Integrate  each  term  separately,  and  write  the  sum  of 
their  integrals  equal  to  an  arbitrary  constant. 

(2)  M  and  N  homogeneous  functions  of  x  and  y  of  the  same 
degree. 

Introduce  in  place  of  y  the  new  variable  v  defined  by 
the  equation  y  =  vx,  and  the  equation  thus  obtained  can 
be  solved  by  (1), 

Or,  multiply  the  equation  through  by ,  and  its 

3fx  +  ^y 
first  member  wiU  become  an  exact  differential,  and  the 
solution  ma}^  be  obtained  by  (6).' 


KEY.  167 

(3)  Of  the  form  (ax  -j-by  +  c) dx  +  (a'x  +  b'y  +  c') dy  =  0. 

If  a6'— a'6  =  0,  the  equation  ma}'  be  thrown  into  the 

a' 
form    {ax  +  by  -{-c) dx  -\-  -  (ax  -\- by -\-c)dy  =  0.      If    now 

z  =  ax-\-  by  be  introduced  in  place  of  either  x  or  ?/,  the  re- 
sulting equation  can  be  solved  by  (1). 

If  ab'—  a'b  does  not  equal  zero,  the  equation  can  be 
made  homogeneous  by  assuming  x  =  x'—  a,  y  =  y'—  ^,  and 
determining  a  and  /3  so  that  the  constant  terms  in  the  new 
values  of  31  and  iV  shall  disappear,  and  it  can  then  be 
solved  by  (2) . 

dv 

(4)  Linear.     General  form  -—-{-  Xiy  =  Xo,  where  X^  and  X2 

are  functions  of  x  alone. 

Solve  on  the  supposition  that  X2  =  0  by  ( 1 )  ;  and  from 
this  solution  obtain  a  A^alue  for  y  involving  of  course  an 
arbitrary-  constant  C.  Substitute  this  value  of  y  in  the 
given  equation,  regarding  C  as  a  variable,  and  there  will 
result  a  differential  equation,  involving  C  and  x,  whose  so- 
lution by  (1)  will  express  C  as  a  function  of  x.  Substitute 
this  value  for  C  in  the  expression  abeady  obtained  for  y, 
and  the  result  will  be  the  requu-ed  solution. 

dv 

(5)  Of  the  form  -^  -f  Xiy  =  X22/",  where  X^  and  X2  are  func- 
tions of  X  alone. 

Divide  through  by  y",  and  then  introduce  z  =  2/^~"  in 
place  of  y,  and  the  equation  will  become  linear  and  may  be 
solved  by  (4). 

(6)  3fdx  +  Ndy  an  exact  differential.     Test  DyM=  D^N. 
Find  I  3Idx^  regarding  y  as  constant,  and  add  an  arbi- 
trary' function  of  y.     Determine  this  function  of  y  by  the 
fact  that  the  differential  of  the  result  just  mentioned,  taken 
on  the  supposition  that  x  is  constant,  must  equal  Ndy. 

Write  equal  to  an  arbitrary  constant  the  |  Mdx  above  men- 
tioned plus  the  function  of  y  just  determined. 


168  INTEGRAL  CALeULUS. 

(7)  Mx  +  Ny=0. 

Multiph'  the  equation  through  by  ,  and  the 

Mx  —  Ny 

first  member  will  become  an  exact  differential.     The  solu- 
tion maj'  then  be  found  b}"  (6). 

(8)  Mx  -  Ny  =  0. 

Multiply  the  equation  throuo-h  by ,   and  the 

Mx  4-  JSfy 

first  member  will  become  an  exact  differential.     The  solu- 
tion may  then  be  found  b}'  (6) . 

(9)  Of  the  form  fi(xy)ydx-^f2{xy)xdy  =  0: 

Multiply  through  by  ,  and  the  first  member 

Mx  —  JSfy 

will  become  an  exact  differential.     The  solution  may  then 
be  found  by  (6). 

(10)  DyM—D^N^  ^  function  of  x  alone. 

Multipl}^  the  equation  through  b}"  e-^         n  ,  and 

the  first  member  will  become  an  exact  difierential.     The 
solution  may  then  be  found  by  (6). 

(11)  D.N-D^M^  ^  fmiction  of  y  alone. 

M  m^N-DyM 

Multiply  the  equation  through  by  e-'        m  ,  and 

the  first  member  will  become  ai^  exact  differential.     The 
solution  maj^  then  be  found  by  (6). 

(12)  DyM-D,N^  ^  function  of  {xy) 


Ny  —  Mx  i-D^M  -  D^N- 


dv 


Multiply  the  equation  through  by  &'  Ny-Mx  where 
V  =  xy^  and  the  first  member  will  become  an  exact  differ- 
ential.    The  solution  may  thus  be  found  by  (6). 

,^3.       x^{D^N-D,M)-\-nNx   ^  ^^^^^^.^^^  ^f.  ?/ .  ^^  ^^.^^^  ^^^. 
Mx+Ny  X 

number. 


KEY.         ■  169 

r  y 

Multiplj'  the  equation  through  hy  aj"ey^"''",  where  ""  =  ^' 
and  fv  =  ^'(-Px^^— ^,^^/)  +  »^^«  and  the  first  member  will 

become  an  exact  difterential.  The  solution  may  then  be 
found  by  (6). 

(14)  Can  be  solved  as  an  algebraic  equation  in  p,  where  p 

stands  for  -^. 
dx 

Solve  as  an  algebraic  equation  in  p,  and,  after  trans- 
posing all  the  terms  to  the  first  member,  express  the  first 
member  as  the  product  of  factors  of  the  first  order  and 
degree.  Write  each  of  these  factors  separatel}'  equal  to 
zero,  and  find  its  solution  in  the  form  V  —  c^O  \>y  (V.). 
Write  the  product  of  the  first  members  of  these  solutions 
equal  to  zero,  using  the  same  arbitrary  constant  in  each. 

(15)  Involves  onl}'  one  of  the  variables  and  p,  where  x>  stands 

for  '^. 
dx 

'Bj  algebraic  solution  express  the  variable  as  an  expli- 
cit function  of  p,  and  then  differentiate  through  relativel}^ 
to  the  other  variable,  regarding  p  as  a  new  variable  and 

remembering  that  —  =  —     There  will  result  a  differen- 
dy     p 

tial  equation  of  the  first  order  and  degi'ee  between  the  sec- 
ond variable  and  p  which  can  be  solved  b}'  (1).  Elimi- 
nate p  between  this  solution  and  the  given  equation,  .and 
the  resulting  equation  will  be  the  required  solution. 

(16)  Of  the  form  xf^p  -\-yf2P  =fzP^  where  p  stands  for  -^. 

Differentiate  the  equation  relative^  to  one  of  the  vari- 
ables, regarding  p  as  a  new  variable,  and,  with  the  aid  of 
the  given  equation,  eliminate  the  other  original  vai'iable. 
There  will  result  a  linear  differential  equation  of  the  first 
order  between  p  and  the  remaining  variable,  which  may  be 

simplified  b}^  striking  out  any  factor  not  containing  -L  or 


170  INTEGEAL   CALCULUS. 

-^,  and  can  be  solved  b^-  (4) .     Eliminate  j?  between  this 

dy 

solution  and  the  given  equation,  and  the  result  will  be  the 

required  solution. 

(17)  Homogeneous  relatively  to  x  and  y. 

Let  y  =  vx,  and  solve  algebraically  relatively  to  p  or  u, 
p  standing  for  -^.     The  result  will  be  of  the  form  p>  —fvi 

or  V  =  Fp.     If  p  =fv,  -  =fv,  -^  =fv,  X-  +  V  =/., 

y 
an  equation  that  can  be  solved  by  (1 ) .     If  ?;  =  Fp^  -  —  Fp^ 

y  =  xFp,  an  equation  that  can  be  solved  by  (16). 

(18)  Of  the  form  i^(<^,i/')  =  0,  where  ^  and  i/^  are  functions  of 
ic,  y  and  — ,  such  that  (^  =  a  and  i/^  =  6  will  lead,  on  differ- 

CtQO  . 

entiation,  to  the  same  differential  equations  of  the  second 

order.  , 

ay 
Eliminate  -f-  between  <ji  =  a  and  il/  =  b,  where  a  and  b 

are  arbitrary  constants  subject  to  the  relation  that 
F(a,h)  =  0,  and  the  result  will  be  the  required  solution. 

(19)  Singular  solution  will  answer. 

Let  ^—p)  and  express  p  as  an  explicit  function  of  x 

and   y.     Take    -i-,    regarding    x    as    constant,    and    see 
^  dy        ^         "^  ' 

whether  it  can  be  made  infinite  b}^  writing  equal  to  zero 
any  expression  involving  y.  If  so,  and  if  the  equation 
thus  formed  will  satisfy-  the  given  diflferential  equation,  it 
is  a  singular  solution. 

Or  take  — ^i-^,  regaixling  y  as  constant,  and  see  whether 
dx 

it  can  be  made  infinite  by  writing  equal  to  zero  an}^  ex- 
pression involving  x.  If  so,  and  if  the  equation  thus 
formed  is  consistent  with  the  given  equation,  it  is  a  singu- 
lar solution. 


KEY.  171 

(20)       Linear,    with   constant   coefficients.      Second   member 
zero. 

Assume  y  =  Ce^"" ;  C  and  m  being  constants,  substitute 
in  the  given  equation,  and  then  divide  through  by  Ce'"^. 
There  will  result  an  algebraic  equation  in  m.  Solve  this 
equation,  and  the  complete  value  of  y  will  consist  of  a 
series  of  terms  characterized  as  follows  :  For  ever}^  dis- 
tinct real  value  of  m  there  will  be  a  term  Ce'"'' ;  for  each 
pair  of  imaginar}^  values,  a  +  6V  — 1,  a  —  &V— 1,  a  term 
Ae'^  cos  bx  +  Be"''  sin  bx ;  each  of  the  coefficients  A,  B,  and 
C  being  an  arbitrar}^  constant,  if  the  root  or  pair  of  roots 
occurs  but  once,  and  an  algebraic  pol3'nomial  in  x  of  the 
(r  —  l)st  degree  with  arbitrary  constant  coefficients,  if  the 
root  or  pair  of  roots  occurs  r  times. 


(21)       Lmear,  with  constant  coefficients.     Second  member  not 
zero. 

Sol¥e,  on  the  h3'pothesis  that  the  second  member  is 
zero,  and  obtain  the  complete  value  of  y  by  (20) .  De- 
noting the  order  of  the  given  equation  hy  w,  form  the  n  —1 

civ  cl^v       ci^    V 

successive  derivatives  — ,  — ^ f^  •    Then  differentiate 

dx  dxr       dx'^ 

y  and  each  of  the  values  just  obtained,  regarding  the  arbi- 
trary constants  as  new  variables,  and  substitute  the  result- 
ing values  in  the  given  equation  ;  and  by  its  aid,  and  the 
«  —  1  equations  of  condition  formed  by  writing  each  of  the 
derivatives  of  the  second  set,  except  the  /ith,  equal  to  the 
derivative  of  the  same  order  in  the  iirst  set,  determine  the 
arbitrary  coefficients  and  substitute  their  values  in  the  ori- 
ginal expression  for  y. 

Or,  if  the  second  member  of  the  given  equation  can  be 
got  rid  of  by  differentiation,  or  by  differentiation  and 
elimination,  between  the  given  and  the  derived  equations, 
solve  the  new  differential  equation  thus  obtained  by  (20) , 
and  determine  the  superfluous  arbitrary  constants  so  that 
the  given  equation  shall  be  satisfied. 


172  INTEGRAL   CALCULUS. 

(22)  Of  the  form  (a  +  bx^p'  +  A(a  +  bxy-^'^^LJ-  +  + 

Ly  =  X,  where  X  is  a  function  of  x  alone. 

Assume  a  +  bx  =  e*,  and  change  the  independent  vari- 
able in  the  given  equation  so  as  to  introduce  t  in  place  of 
X.     The  solution  can  then  be  obtained  by  (21). 

(23)  Either  of  the  primitive  variables  wanting. 

Assume  z  equal  to  the  derivative  of  lowest  order  in  the 
equation,  and  express  the  equation  in  tei'ms  of  z  and  its 
derivatives  with  respect  to  the  primitive  variable  actually 
present,  and  the  order  of  the  resulting  equation  will  be 
lower  than  that  of  the  given  one. 

(24)  Of  the  form  ---^=  X.     X  being  a  function  of  x  alone. 
^     ^  da;" 

Solve  b}'  integrating  n  times  successively  with  regard 

to  X. 

Or  solve  by  (21). 

d'li 

(25)  Of  the  form  y4=  ^-      ^  being  a  function  of  y  alone. 


Multiply  by  2--^  and  integrate  relativelj-  to  x.     There 
dx 


will    result    the    equation   f  -^i  J  =  2  j  Tdy  +  C,    whence 
—  =  (2  j  Ydy -\-  C*)-^,  an  equation  that  may  be  solved  hy 

d'hi  d"-~^ii 

(26)       Of  the  form  ^  =/y^i- 

d^'-hj         ^.       dz       .         ,        dz  rdz     „ 

Assume  - — ~  =  z,  then  —  =fz  or  dx  =  —,  a;  =  I  —  +  C. 
da;™-^        '  da;     -^  Jz         J  fz 

After  effecting  this  integi'ation,  express  z  in  terms  of  a; 

and  C.     Then,  since  z= ?,  1!  =F(x,C),  an  equa- 

da;""^    dx"~^ 

tion  that  maj'  be  treated  by  (24). 

Or,  smoe  — ^  =  .,  -^,  =  j  M.  +  o  =  J  -  +  c,  smce 


KEY.  173 

zdz 


-=|  i3=/-(/f-)-'=/l(/7?-, 

+  Ci, Continue  this  process  until  ?/ is  expressed  in 

terms  of  z  and  71  —1,  arbitrary  constants,  and  then  elimi- 

Cdz 
nate  z  hj  the  aid  of  the  equation  x=  I  — -  +  0. 


(27)  Of  the  form  ^^  =/^^f  • 

Let ^  =  2^,  and  the  equation  becomes  -r—,  =fz,  and 

ma}'  be  solved  b^'  (25). 

(28)  Homogeneous  on  the  supposition  that  x  and  y  are  of  the 

degree  1,  _j/  of  the  degree  0,  —4  of  the  degree  —1, 

dx  dx- 

Assxime  .^•=e^,  y  =  e'^z,  and  bj'  changing  the  variables 
introduce  6  and  z  into  the  equation  in  the  place  of  x  and  y. 
Divide  through  by  e^  and  there  will  result  an  equation  in- 
volving only  z,  — ,  ——,,  ,  whose  order  ma}' be  depressed 

du   dd' 
by  (23). 

(29)  Homogeneous  on  the  supposition  that  x  is  of  the  degree 

1,  1/  of  the  degree  n,  -^  of  the  degree  ?^  —  1,  -^  of  the  de- 

'  -^                 ^  dx  °  dx^ 

gree  n  —  2, 

Assume  x  =  e^,  y  =  e"-^z,  and  by  changing  the  variables 

introduce  6  and  z  into  the  equation  in  the  place  of  x  and  y. 

The  resulting  equation  may  be  freed  from  0  by  division 

and  treated  b}'  (23). 

(30)  Homogeneous  relativeh'  to  y,  — ,  — f , 

dx   dx- 

Assume  y  =  e",  and  substitute  in  the  given  equation. 
Divide  through  b}'  e^  and  treat  by  (23). 


(31)       Containing  the  first  power  only  of  the  derivative  of  the 
highest  order. 

The  first  member  of  the  equation  may  be  an  exact  de- 
rivative ;  call  it If  w  is  the  order  of  the  equation, 

dx 


174  INTEGRAL   CALCULUS. 

represent  ^^^-^  by  p  and  LI  by  ^.     Multiply  the  term 
f/a-""^  cto"        ax 

containing  —  b}-  dx  and  integrate  it  as  if  p  were  the  only 

variable,  calling  the  result  Ui ;  then  replacing  p  by f , 

find  the  complete  derivative  -y-,  and  form  the  expression 

— — -i,  representing  it  by  — ^.    If  — ^  contains  the  first 

dx       dx  dx  dx 

power  only  of  the  highest  derivative  of  y,  it  may  itself  be  an 
exact  derivative,  and  is  to  be  treated  precise^  as  the  first 

dV 
member  of  the  given  equation  —  has  been.    Continue  this 

dx 

dV 
process  until  a  remainder  — ^^^  of  the  first  order  occurs. 

dx 

Write  this  equal  to  zero,  and  solve  b}'  (V,),  throwing  its 
solution  into  the  form  F„_i  =  C.    A  complete  first  integral 

of  the  given  equation  will  be  C/i  +  ?72  + +  Vn^x  =  C. 

The  occurrence  at  any  step  of  the  process  of  a  remainder 

— -*,  containing  a  higher  power  than  the  first  of  its  highest 
dx 

derivative  of  y,  shows  that  the  fii'st  member  of  the  given 
equation  was  not  an  exact  derivative,  and  that  this  method 
will  not  apply. 


(32)       Oftheform^+X^  +  1 


=  0,  where  X  is  a 


cly 
dx 
function  of  x  alone  and  Ya  function  of  y  alone.    Multiply 


through  by 


'dy 
dx 


and  the  equation  will  become  exact,  and 


may  be  solved  by  (31). 
(33)       Singular  integral  will  answer. 

fln-ly  cZ"?/  do 

Call  ^^  p,  and  ^  q,  and  find  ^^,  regarding  j?  and  q 

as  the  only  variables,  and  see  whether  ^  can  be  made 

•^  dp 

infinite  by  writing  equal  to  zero  any  factor  containing  p. 


KEY.  175 

If  SO,  eliminate  q  between  this  equation  and  the  given 
equation,  and  if  the  result  is  a  solution  it  will  be  a  singular 
integral. 

(34)       General  form,  Pdx  +  Qdy  +  Rdz  =  0. 

If  the  equation  can  be  reduced  to  the  form  Xdx  +  Ydy 
-\-  Zdz  =  0,  where  X  is  a  function  of  x  alone,  Y  a  function 
of  y  alone,  and  Z  a  function  of  z  alone,  integrate  each 
term  separatel}',  and  write  the  sum  of  tlie  integrals  equal 
to  an  arbitrary  constant. 

If  not,  integrate  the  equation  by  (V.)  on  the  supposition 
that  one  of  the  variables  is  constant  and  its  differential 
zero,  writing  an  arbitrary  function  of  that  variable  in  place 
of  the  arbitrary-  constant  in  the  result.  Transpose  all  the 
terms  to  the  first  member,  and  then  take  its  complete 
differential,  regarding  all  the  original  variables  as  variable, 
and  write  it  equal  to  the  first  member  of  the  given  equa- 
tion, and  from  this  equation  of  condition  determine  the 
arbitrary  function.  Substitute  for  the  arbitrar}'  function 
in  the  first  integral  its  value  thus  determined,  and  the 
result  will  be  the  solution  required. 

If  the  equation  of  condition  contains  any  other  varia- 
bles than  the  one  involved  in  the  ai-bitrar}"  function,  they 
must  be  eliminated  b}'  the  aid  of  the  primitive  equation 
already  obtained ;  and  if  this  elimination  cannot  be  per- 
formed, the  given  equation  is  not  derivable  from  a  single 
primitive  equation,  but  must  have  come  from  two  simul- 
taneous primitive  equations. 

In  that  case,  assume  an}^  arbitrary'  equation /( .t;,?/, 2;)  =0 
as  one  primitive,  differentiate  it,  and  eliminate  between  it 
its  derived  equation  and  the  given  equation,  one  variable, 
and  its  differential.  There  will  result  a  differential  equa- 
tion containing  onl}-  two  variables,  which  may  be  solved 
by  (III.),  and  will  lead  to  the  second  primitive  of  the 
given  equation. 


176  LNTEGEAL   CALCULUS. 

(35)  General  form,  Pdx^  +  Qdx2  +  Rdx^  + =  0. 

If  the  equation  can  be  reduced  to  the  form  XidcCi-f-Xadajg 

+  Xgdx^  + =  0,  where  Xi  is  a  function  of  Xi  alone,  X2 

a  function  of  x's  alone,  Xg  a  function  of  x^  alone,  etc.,  inte- 
grate each  term  separatel}',  and  write  the  sum  of  their 
integrals  equal  to  an  arbitrary'  constant. 

If  not,  integrate  the  equation  by  (V.),  on  the  supposi- 
tion that  all  the  variables  but  two  are  constant  and  their 
differentials  zero,  writing  an  arbitrary  function  of  these 
variables  in  place  of  the  arbitrary-  constant  in  the  result. 
Transpose  all  the  terms  to  the  first  member,  and  then 
take  its  complete  differential,  regarding  all  of  the  .original 
variables  as  variable,  and  write  it  equal  to  the  first  mem- 
ber of  the  given  equation,  and  from  this  equation  of  con- 
dition determine  the  arbitrary  function.  Substitute  for 
the  arbitrary  function  in  the  first  integral  its  value  thus 
determined,  and  the  result  will  be  the  solution  required. 

If  the  equation  of  condition  cannot,  even  with  the  aid 
of  the  primitive  equation  first  obtained,  be  thrown  into  a 
form  where  the  complete  differential  of  the  arbitrary  func- 
tion IS  given  equal  to  an  exact  differential,  the  function 
cannot  be  determined,  and  the  given  equation  is  not  deriv- 
able from  a  single  primitive  equation. 

(36)  S3'stem  of  simultaneous  equations  of  the  first  order. 

If  an}'  of  the  equations  of  the  set  can  be  integrated 
separately  by  (II.)  so  as  to  lead  to  single  primitives,  the 
problem  can  be  simplified ;  for  b}'  the  aid  of  these  primi- 
tives a  number  of  variables  equal  to  the  number  of  solved 
equations  can  be  eliminated  from  the  remaining  equations 
of  the  series,  and  there  will  be  formed  a  simpler  set  of 
simultaneous  equations  whose  primitives,  together  with  the 
primitives  already  found,  will  form  the  'primitive  s^'stem 
of  the  given  equations. 

There  must  be  n  equations  connecting  n  +  1  variables, 
in  order  that  the  system  may  be  determinate. 

Let  a;,  x^,  Xo, ,  x^  be  the  original  variables.     Choose 


KEY.  177 

any  two,  x  and  x-^,  as  the  independent  and  the  principal  de- 
pendent variable,  and  by  successive  eliminations  form  the 

n  equations  — i  =f^{xay,X2, , x^) ,  — i  =f%{x,x-i^,X2, , a*„), 

ax  ax 

dv 
,  up  to  ^-^z=f„{^,Xi,X2, ,Xn).     Differentiate  the  first 

of  these  with  respect  to  x  n  —  1  times,  substituting  for 

— -,  — ?, , — ?,  after  each  step  then- values  in  terms  of 

dx     dx  dx 

the  original  variables.  There  will  result  n  equations, 
which  will  express  each  of  the  n  successive  derivatives 

tt»^i  Ct    SO-i        CI    X-i  Ct     Oji         'J  n 

—^    ^'   -W'  '  -7-T'    "^    *®^'™^    O^  ^'    ^1'  ^'2'    '    ^«- 

dx      dx-     dx^  dx"- 

Eliminate  from  these  all  the  variables  except  x  and  Xi, 
obtaining  a  single  equation  of  the  ?ith  order  between  x 
and  Xi.  Solve  this  by  (VII.),  and  so  get  a  value  of  x^  in 
terms  of  x  and  ii  arbitrary  constants.     Find  hj  differen- 

tiatmg  this  result  values  for  — -%  — /, , -/,  and  write 

dx     dxr  dx"  ^ 

them  equal  to  the  ones  already  obtained  for  them  in  terms 
of  the  original  variables.  The  n—1  equations  thus  formed, 
together  with  the  equation  expressing  x^  in  terms  of  x  and 
arbitrary  constants,  are  the  complete  primitive  S3'stem 
required. 

(37)  S5'stem  of  simultaneous  equations  not  of  the  first  order. 
Eegard   each   derivative   of   each   dependent  variable, 

from  the  first  to  the  next  to  the  highest  as  a  new  variable, 
and  the  given  equations,  together  with  the  equations  de- 
fining these  new  variables,  will  form  a  S3'stem  of  simulta- 
neous equations  of  the  first  order  which  ma}'  be  solved  by 
(36).  Eliminate  the  new  variables  representing  the 
various  derivatives  from  the  equations  of  the  solution,  and 
the  equations  obtained  will  be  the  complete  primitive  sys- 
tem requu'ed. 

(38)  All  the  partial  derivatives  taken  with  respect  to  one  of 
the  independent  variables. 


178  INTEGRAL   CALCULUS. 

Integi-ate  hy  (II.)  as  if  that  one  were  the  011I3-  mdepen- 
dent  variable,  replacing  each  arbitrarj-  constant  b}'  an 
arbitrary-  function  of  the  other  independent  variables. 

(39)  Of  the  first  order  and  linear,  containing  three  variables. 
General  form,  PD^z  +  QD^z  =  E. 

Form  the  auxiliary  s^'stem  of  ordinary  differential  equa- 

t/t*       elf/       Ct'^ 
tions  —  =  -^  =  -^,  and  integrate  by  (36).     Express  their 

primitives  in  the  form  u  =  a,  v  =  b,  a  and  b  being  arbi- 
trar}-  constants  ;  and  u—fv^  where  J  is  an  arbitrar}^  func- 
tion, will  be  the  required  solution. 

(40)  Of  the  first  order  and  linear,  containing  more  than  three 

variables.      General   form,    PiDx^z  +  P^Dx^z -\- =  B, 

where  :ci,  x'g,  ,  x„.  are  the  independent  and  z  the  depen- 
dent variables. 

Form  the  auxiliary'  sj'stem  of  ordinary  differential  equa- 
tions ~  —  — ^ =  — 2  =.  _,  and  integTate  them  b}'  (36). 

Express  their  primitives  in  the  form  Vi  =  a,  v,  =  &,  V3  =  c, 
,  and  Vi—f{v2,V3, ,-«„),  where/  is  an  arbitrary  func- 
tion, will  be  the  required  solution. 

(41)  Of  the  first  order  and  not  linear,  containing  three  varia- 
bles, F(x,y,z.2),q)  —  0,  where  p  =  D^s,  q  =  DyZ. 

Express  q  in  terms  of  x,  y,  z  and  p  from  the  given  equa- 
tion, and  snbstitute  its  value  thus  obtained  in  the  auxil- 
iary sj'stem  of  ordinary  differential  equations  — ^- —  =  dy 

d/Z  (Hi)  ^ 

=  =  — — -^ — .     Deduce  by  integration  from 

q  —  pDpq      D,q+pD,q 

these  equations,  b}'  (36),  a  value  of  p  involving  an  arbi- 

trar}'  constant,  and  substitute  it  with  the  corresponding 

value  of  q  in  the  equation  dz  =  pdx  +  qdy.      Integrate 

this  result  by  (34) ,  if  possible  ;  and  if  a  single  primitive 

equation  be  obtained,  it  will  be  a  complete  primitive  of  the 

given  equation. 


ELEY.  179 

A  singular  solution  may  be  obtained  by  finding  the 
partial  derivatives  DpZ  and  Z>j2  from  the  given  equation, 
writing  them  separately  equal  to  zero,  and  eliminating  p 
and  q  between  them  and  the  given  equation. 

(42)  Of  the  first  order  and  not  linear,  containing  more  than 

three  variables.    F{xi^x.2, ,  a;„,  z^pi^p^-t >i>«)  =  0,  where 

Pi  =  Dx^z,  2h  =  Dx^z, 

Form  the  linear  partial  differential  equation  %i\^{DxiF 
+  PiD^F)Dp^^  -  Dp.F{Dx,^  +  PiD^^)'\  =  0,  where  <l>  is 

an  unknown  function  of  (.x'l,  ,  .t,i,  Pi, iP„),  and  where 

%  means  the  sum  of  all  the  terms  of  the  given  form  that 
can  be  obtained  by  giving  i  successively  the  values  1,  2, 
3,  ,  n. 

Form,  by  (40),  its  auxiliary  sj'stem  of  ordinary  differen- 
tial equations,  and  from  them  get,  by  (36),  n  —  1  uite- 

grals,  $1  =  «!,  <I>o  =  Oo, ,  $„_!  =  a„_i.    By  these  equations 

and  tlie  given  equation  express  pi,  p2-,  ,  Pn  in  terms  of 

the  original  variables,  and  substitute  their  values  in  the 

equation  clz  —jhdxi  -{-jhdx.^  + +2J„da;,j.     Integrate  this 

by  (35),  and  the  result  will  be  the  required  complete  primi- 
tive. 

(43)  Of  the  second  order  and  containing  the  derivatives  of 
the  second  order  onl}'  in  the  first  degree.  General  form, 
BD^-z  +  SD^D^z  +  TZ>/z  =  F,  where  i?,  S,  T,  and  Fmay 
be  functions  of  a%  y,  2;,  D^z,  and  DyZ. 

Call  D^z  p  and  DyZ  q. 
Form  first  the  equation 

Rdf  -  Sdxcly  +  Td^  =  0,  [  1  ] 

and  resolve  it,  supposing  the  first  number  not  a  complete 
square,  into  two  equations  of  the  form 

dy  —  7?ii  dx  =  0 ,         dy  —  m^  dec  =  0 .  [2] 

From  the  first  of  these,  and  fi'om  the  equation 

Rdpdy  +  Tdqdx  —  Vdxdy  =  0,  [3] 


180  INTEGRAL   CALCULUS. 

combined  if  needful  witli  the  equation 

dz  =  2)c?a;  +  qdy, 

seek  to  obtain  two  integrals  Ui  =  a,  Vj  =  /3.  Proceed- 
ing in  the  same  way  with  the  second  equation  of  [2], 
seek  two  other  integrals  Ug  =  ai,  Vj  =  /Si ;  then  the  first  in- 
tegrals of  the  proposed  equation  will  be 

'ih=fiV,         ih=f2V2,  [4] 

where /i  and/2  denote  arbitrary  functions. 

To  deduce  the  final  integral,  we  must  either  integrate 
one  of  these,  or,  determining  from  the  two  p  and  q  in  terms 
of  X,  y,  and  z,  substitute  those  values  in  the  equation 

which  will  then  become  mtegrabie.  Its  solution  will  give 
the  final  integral  sought. 

If  the  values  of  nii  and  wi2  are  equal,  only  one  first  in- 
tegral will  be  obtained,  and  the  final  solution  must  be 
sought  b}'  its  integration. 

When  it  is  not  possible  so  to  combine  the  auxiliary 
equations  as  to  obtain  two  auxiliary  integrals  u  =  a,  v  =(3, 
no  first  integral  of  the  proposed  equation  exists,  and  this 
method  of  solution  fails. 


Examples. 

(1)  sin,:«  cos  ydx  —  coscc  sinydy  =  0.  Ans.  cosy  =  c  cos  a;. 

(2)  [2  a/ (xy)  —  ic]  dy  +  ydx  =  0.  Ans.  y  =  ce~  V^. 

(3)  {2x-y  +  l)dx  +  {2y-x-l)dy  =  0. 

Ans.  x^  —  xy  +  y^  -\-  X  —  y  =  c. 

(4)  ^4-ycosx  =  5H^.  Ans.  ?/  =  sina;  -  1  +  ce"""'. 
dx                          2 


KEY. 


181 


(5)   (l-af)^-xy  =  axyK        Ans.  y  =  lc^{l -x-)-ay\ 


dx 


xdy  —  ydx      „  .   „    ^ +  2/"  ,  4-„,,-iy  _ 

(6)  xdx  +  2/d2/  +     i;,  ,.    =  0.       ylns.  —2—  +  tan    -  _ 


a^  +  2/^  '  2 


w^=4M-(l)- 


^47is.  y  =  ex  +  c  —  (?. 

Singular  solution,  y  =  ^^^— 

4 


(8)   /'^Y_£!  =  0.     Ans.   (y-a\ogx-c)(y+alogx-c)  =  0. 
\dxj      X- 

^  ^   doj*         da^^    dx^         dx      -^ 

Ans.  ?/  =  (co  +  Cia;  +  C2a3-  +  C8a^)€^ 

(10)  pi  -2k'M  +  ^-22/  =  e\      Ans.  y  =  (ci+C2o;)e^^+  — ^- 
dx-  dx  K'^  —  ^) 

(11)  ^-|_i^==o.  ^ws.  2/ =  c  log  a;  +  c'. 
die-      a;  dx 

(12)  a;2^'  +  a;^^+(2x-?/-l)^+?/2  =  0.  Find  a  first  integi-al. 

da;3         dx"  dx  -.o  .        .7,. 

dxr         dx 

(13)  _^  +  _^  +  -^=0.     Ans.   {x-a){y-b){z-c)=C. 
x—a      y  —  h      z—G 

(14)  (y-{-z)dx-i-dy  +  dz  =  0.  Ans.  e'{y  +  z)  =  c.    ' 


(15)  ^'  +  4a;  +  ^=0,     ^i^  +  3?/-a^  =  0. 
d<  4  ai 


Ans.  x  =  ce2_^,      2/  =  (c^+ ^j)  e  2. 

(16)^4-771^0^=0,     ^-m2a;  =  0.         Ans. 
^     '    dt'  df 

riT^  Dz  =  — ^^*«-  e'^a;  +  2/ +  2)  =  <^y. 

^     ^       ^        a;  +  2; 


182  INTEGEAL   CALCULUS. 


(18)  xzD^z-\-yzDyZ  =  xy.  Ans.  z^  =  xy -\- <j)(z\. 

(19)  D^z.DyZ=l.  Ans.  z=ax  +  -  +  b. 

(20)  x'D,'z  +  2xyD,DyZ+f-Dy'z  =  0.     Ans.  z=xcf,(^  ^^i^A 

(21)  {DyzYD^z  -  2I)^zDyZD^DyZ  +  {D^zyD^'z  =  0. 

Ans.    y  =  X(^z  +  \\/z. 

(22)  D^z.D^DyZ  — DyZ.D^'z  =  0.  Ans.    x=(j)y  +  ij/z. 


APPENDIX. 


Chap.  V.]  INTEGEATION.  65 


CHAPTER  V. 

INTEGEATION. 

74.  We  are  now  able  to  extend  materially  our  list  of  formulas 
for  direct  integration  (Art.  55) ,  one  of  which  may  be  obtained 
from  each  of  the  derivative  formulas  in  our  last  chapter.  The 
follo\\ang  set  contains  the  most  important  of  these  :  — 

Dj. log x=  -  gi ve s  X: -  =  log x . 

X  X 

D^a'  =  a^loga  "     /^  a"^  log  a  =  a"" . 

D^e'^e'  "     Xe^  =  e^ 

D^  sin  X  =  cos  x  "     /x  cos  x  —  sin  x. 

D^  cos  X'  =  —  sin  ic  "     X  ( —  sin  x)  =  cos  x. 

D^  log  sin  X  =  ctn  x  "X  ctn  x  =  log  sin  x. 


D^logcosx  =  —  tana; 

(( 

fx{—  tan X")  =  log  cos x. 

7")     oin — 1  /^»  

C( 

/-          1               •  -1 

L =  sm   ^x. 

V(i-a^) 

7~)  foTi — ^^ 

ii 

X      ^    ,-tan-^x. 
1  +ir 

7^      T70T>0 1  A»    

li 

r              1                                1 

/, —  =  vers  ^x. 

JL-'-j.   V  CI  O          iO  >— — 

^{2x-x^) 

The  second,  fifth,  and  seventh  in  the  second  group  can  be 
written  in  the  more  convenient  forms, 


66  DIFFERENTIAL   CALCULUS.  [Art.  75. 

log  a 
y^sina;=  —  cosa;; 
f^tsLUX—  —  logcosaj. 

75.  When  the  expression  to  be  integrated  does  not  come  under 
an}^  of  the  forms  in  the  preceding  list,  it  can  often  he  iwepared 
for  integration  by  a  suitable  change  of  variable,  the  new  variable, 
of  course,  being  a  function  of  the  old.  This  method  is  called 
integration  by  substitution,  and  is  based  upon  a  formula  easily 

deduced  from  D^(Fy)=DyFy  .D^y ; 

which  gives  immediately 

Fy=A{D,Fy.D^y). 


Let 

u=D,Fy, 

then 

Fy=fy^h 

and  we  have 

fyU=f^{uD^y); 

or,  interchanging  a;  and  y, 

fu=^f^{uD,x).  [1] 

For  example,  required       f{a  +  6x)". 

Let  '  z  =  a  +  bx, 

and  then  X(a  +  bxy  =/,z-  =f,{z^ .  D,x) ,  by  [1]  ; 

,    ,  z      a 

but  x= , 

b      b 

B.x  =  - ; 

0 

hence  /,(a  +  bxY  =  {f^z'^=l  ^"^ 


b  bn  +  1 


Chap.  V.]  INTEGRATION. 

Substituting  for  z  its  value,  we  have 

_1  (a  +  5a;)»+^ 


67 


Jl{a  +  hx) 


b       71+1 
Example. 


Find  X 


a  +  bx 


Ans.    -log (a  +  bx) . 


76.    Iffx  represents  a  function  that  can  be  integrated, /( a  +  bx) 
can  always  be  integi'ated  ;  for,  if 


then 
and 

Find 

(1)  f^sinax. 

(2)  f^cosax. 

(3)  y^tanaa". 

(4)  7^  ctn  ace. 

77.    Required/^ 

L 


z=  a  +  bx, 
D,x  =  - 


fj{a  +  6a;)  =fjz  =fJzD,x  =  ^^fjz. 


Examples. 


Ans. cosaa;. 

a 

Ans.   —sin  ace. 
a 


1 


1 


yj  {a-  —  X-)      a 


i^-m 


Let 
then 


X 
Z  =  -, 

a 
x  =  az, 

D^x  =  a, 


68 


DrETEEENTIAl.  CAliCULTIS. 


1/. 

a 


L 


=  i/. 


a-'VCl-^')       ci'^il-z") 


[Art.  78. 


1 


Examples. 


=  sin~^2  =  sin~^  -. 


Find 


(^\    r 

1              X 
Ans.    -tan~^-' 

^^^   -Sr+ct-^ 

(^\   r                   . 

X 

Ans.   vers  ^-' 

C6 

^"^   -^VC^a^-^O 

78.    Required  f^—-—, -• 

V(.^"  +  «-)     ■ 

Let                                  z  =  x  +  ■sj(x^  +  a^)  ; 

then                                 z  —  x  =  ^{a?  +  a^) , 

z^  —  2  zx  -i-  (kP  =  x^  -\-  a^, 

2zx  =  z-  —  a^, 

0.  =  ^'-^', 
2z    ' 

2;2  _  0,2 

z'^  +  ci? 

2z 


2z 


D^x 


z-  +  or 
2  2- 


/. 


=X^,=/.-^^^^^^ 


^  ( .^•^  +  a-)        "^  2;-  +  a-         2^  +  a^ 


=/. 


2z     z^-\-a?   '  ~\  ,      ^  /-5 •„, 

72  "ir:^  =•/--  =  log2  =  log(;«  +  ^xr  +  a-) . 


z^  +  ct^     2  2;^ 


Example. 


FindX 


sji.^-o?) 


Ans.   log  {x  +  Va/*^  —  a-) , 


Chap.  V.]  INTEGRATION.  69 

79.    When  the  expression  to  he  integrated  can  he  factored,  the 
required  integi'al  can  often  be  obtained  by  the  use  of  a  formula 

deduced  from  D^{iiv)  =  uD^v  +  vD^u, 

which  gives  uv  =  f^  uD^  v  -\-f^  vD^  u 

or  f^uD^v  =  uv —f^vD^u.  [1] 

This  method  is  called  integrating  hy  parts. 

(a)    For  example,  required  f\ogx. 

\ogx  can  be  regarded  as  the  product  of  logic  by  1. 

Call  log  a;  =  u  and  1  =  D^v, 

then  D^u  =  -, 

X 

v  =  x; 
and  we  have 

^logcc  =/^l  logo;  =f^uD^v  =  uv  —f^vD^u 


=  a;  log  a;  —fx-  =  ^loga;  —  x. 

X 


Example. 
Find  y^O/*  log  cc. 
Suggestion :  Let    loga;  =  u  and  x  =  D^v. 


Ans.    -  x^  ( loffoj 


80.    Required fsin^x. 
Let  u  =  sin  a?  and  D^v  =  sin  a;, 

then  D^it— cos  X, 

v=  —  cosa;, 
7^sin^a;=  —  sina^cosa;  +/^cos^a; ; 


70  DIFFEBENTIAIi  CALCUI.US.  [Art.  81. 

but  cos^x  =  1  —  srn^x, 

so  Xceos^x=f^l—f^sm-x  =  x—f^sm^x 

and  /x  sin-  x  =  x  —  sin  x  cos  x  —  f^  sin^  a;. 

2Xsin-a'  =  x  —  sin  .'c  cos  a?. 
f^  sin^  x=  ^(x  —  sin  x  cos  x) . 

Examples. 

(1)  FindXcos^a;.  Ans.   -(a;  +  sina;cos.T). 

(2)  /j-sinxcoscc.  Ans.    -. 

81.    Very  often  both  metliods  described  above  are  required  in 
the  same  integration. 
(a)    Required f^sm~'^x. 


Let 

sin~\T  =  ?/, 

then 

x=  sin?/ ; 

DyX=  cosy, 

f,sm-^x  =f^y  =f^ycosy. 

Let 

u  =  y  and  DyV  =  cosy  ; 

then 

A«  =  l, 

V  =  sin?/, 

and 

/j,  ?/ cos  2/ = ?/ sin  ?/ — /j,  sin  ?/ =  ?/ sin  ?/ -|- cos  ?/ = a;  sin"  ^T  +  ^y  ( 1  —  .1'-) . 

Any  inverse  or  anti-function  can  be  integrated  by  this  method 
if  the  direct  function  is  integi-able. 

(6)    Thus,  fJ-'x=f^y=f^yDJy=:yfy-fJy 

where  y=f-^x. 


Chap.  V.]  INTEGRATION.  71 

Examples. 

(1)  Find/a.cos~^a;.  Ans.   a;cos~^cc  — ^(1  —  x^). 

(2)  f^iari~^x.  A71S.    a;tan~^a;  — -log(l  +  a^). 

(3)  yivers~^£c.  Ans.    {x—l)yevs~^x-{-^(2x—x^). 

82.    Sometimes  an  algebraic  transformation,  either  alone  or  in 
combination  tvith  the  preceding  methods,  is  useful. 

1 


(a)    Required/^ 


af—  a- 


x^—  a'      2 a  \x  —  a      x  +  aj 
and,  by  Art.  75  (Ex.), 

A i  =  ^  Dog(^  -  a)  -  log(.«  +  a)  ]  =  —  log'^^. 

XT— a'      2  a  2a       x-i-a 

(b)    Required/,    \(:^^. 

\\i-xj    V(i-^')     VCi-^")     vci-^-" 


/x 


1 


.  —  oin      -*^o' 


fx — — ^ 5-  can  be  readily  obtained  by  substituting  ?/  =  (1  —  a^, 

^(1  — ar) 

and  is  —  ^/(l  — a.*^)  ; 

hence  /     /('l±|')  =  sin-^o;  _  ^(1  _  ^2)  _ 

(c)    Required  f^-^/  (a^  —  x^) . 


V(a'-a^)  = 


V  (a^  -  x')       V  («'  -  ^')       V  (a'  -  »^) ' 


72  DIFFERENTIAL   CALCULUS.  [Art.  83. 

and         /,V(a^  -  ^'^)  =  a^f.    „  }      ,,  -/» 


whence        /,  V («^  -  ic^)  =  a^  sin"^  ^  -/^    ,,  f" — 5-,  by  Art.  77  ; 


af 


by  integration  by  parts,  if  we  let 

w  =  V  (^^^  ~  ^^)  ^iicl  Z)^  V  =  1. 
Adding  our  two  equations,  we  have 

a 
and  .  • .  X  V  (a'  -  a^)  =  |  Ma^-x"  +  a^  sin"^  ^]  • 

Examples. 
Find 

(1)  /.^{x'  +  a'). 

(2)  /,V(a^-^-a^). 

^ws-    ^  [a; V (^  -  a')  -  aHog(x  +  Va;-  -  a^) ] . 


Applications. 

83.    To  find  the  area  of  a  segment  of  a  circle. 
Let  the  equation  of  the  circle  be 

ar  +  2/"  =  "^  J 

and  let  the  required  segment  be  cut  off  by  the  double  ordinates 
thi'ough  {xQ,y^  and  {x,y) .     Then  the  requu-ed  area 

A=2f^y  +  C. 


Chap.  V.] 


INTEGE.ATION. 


73 


From  the  equation  of  the  circle, 

hence  A  =  2/^  V  («^  -x-)  +  C; 

and  therefore,  b^-  Art.  82  (c) , 

A  =  x-sJia'  -  x")  +  a'  sin"^  |  +  C. 

As  the  area  is  measiu'ed  from  the  ordinate  i/q  to  the  ordinate  y^ 
^  =  0  when  X  —  Xq  ; 

therefore  0  =  Xq  V  («  —  -V)  +  «"  sin~  ^ h  C*, 


C  =  —  .To  V (f''  —  ^'o')  —  «"  sill  ^ — ' 
and  "we  have 

^  t=  aj  ^  (a^  —  it'-)  +  a^  sin~-^ ct'o  V  ("^'^ — ^"o^)  -~  «'  sin~ ^ — 

If  0^0  =  0,  and  the  segment  begins  luith  the  axis  ofY, 

X 

A  =  X'^  (a^  —  ar)  +  «^  sin~^— • 
If,  at  the  same  time,  x  =  a,  the  segment  becomes  a  semicircle,  and 

A=  a  ^/(a^  —  a^)  +  a^sin"^-  =  ~ — 

ct        ^ 

The  area  of  the  whole  circle  is  -a^. 


74 


DEFFEEENTIAIi   CALCULUS. 


[Art.  84. 


EXAIVIPLES.. 

(1)    Show  that,  hi  the  case  of  an  ellipse, 


.1=' 


£:+iC=i, 

a-       h- 


the  area  of  a  segment  beginning  with  any  ordinate  y^  is 

X  -Jia^  —  it"')  +  a^sin"^^ Xf^JCa^—  Xq)  -^  a^sin" 

a 

That  if  the  segment  begins  with  the  minor  axis, 

^=-     ccy(«-  — ar)  + trsm     - 
a  I  a 

That  the  area  of  the  whole  ellipse  is  -ab. 

(2)    The  area  of  a  segment  of  the  hj^^erbola 

cr       b' 


A  =  x-sJ(x^  ~  a-)  —  fr  log(cc  +  Vic^  —  «') 


—  Xo  V(^'o"  —  «")  +  «"log(a;o  +  Va^o^  —  a-). 
If  a'o  =  a,  and  the  segment  begins  at  the  vertex, 


A  =  X'sJ{a^  —  cr)  —  a- log  (a;  +  V-»2  —  ^^2)  +  a'logrt. 

84.    To  find  the  length  of  any  arc  of  a  circle,  the  coordinates 
of  its  extremities  being  (cccyo)  ^^^^  i^iV)  • 

By  Art.  52,  s=/.V[l+ (A^)']- 

From  the  equation  of  tlie  circle, 

x^  +  y-=  a% 


Chap.  V.J  INTEGRATIOlSr.  75 

we  have  2x-\-2yD^y  =  0, 

y 

r       f 

s=f^^=af^        }  =asin-^^>a    (Art.  77.) 

When  cc  =  i»07  s  =  0  ; 

hence  0  =  a  sin~  ^  -  +  C, 

C  =  —  asm  ^-, 
a' 

.  /^  .   _i^         .      i-'^o 

and  s  =  a   sin   ^ sin~^  — 

\         a  a 

Ka;o=  0,  and  ^7ie  arc  is  measured  from  the  highest  point  of  the 

X 

c  ircle ,  s  =  a  sin~  ^  -  • 

'  a 

If  the  arc  is  a  quadrant,     x  =  a, 

•  _i/i\      T^a 
s  =  asin  1(1)  =  —  , 

and  the  whole  circumference  =  2-a. 


85.    To  find  the  length  of  an  arc  of  the  parabola  y^=  2mx. 
We  have  2yD^y  =  2m  ; 

7-v  m 

Ay  =  — ; 
y 


76 


DIFFEEENTIAL   CALCULUS. 


[Art.  85. 


5=/. 


J 


'sl{m?  +  y'^) 


=  fy 


~^(m^  +  y'^)  DyX 


by  Art.  73  ; 


s  =  —fy  ^/mF+f=—-ly\/7n^  +  y^-+  mHogiy  +  Vm-+  ?/-)]  +  C, 
m  2m 

by  Art.  82,  Ex.  1. 
If  the  arc  is  measured  from  the  vertex, 

s  =  0  when  y  =  0; 


0  = (m^  log  m)  +  0^ 

2  m 


C= mlogm, 


and 


s  = 


'y-^jim'  +  y^)  y  +  ^  (rri^ -^  y-)' 
+mlog- 


m 


Example. 


Find  the  length  of  the  arc  of  the  curve  a^=  27 ^/^  included  be- 
tween the  origin  and  the  point  whose  abscissa  is  15. 

Ans.    19. 


Da*' 


r 


BOSTON  COLLEGE 


'O^ 


i      3  9031   01548937  0 


ON  C0U£GF5!Mr 


^60413 


DOES  NOT  CiRCULAi 


B^^ 


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